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## Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

Access the Lesson-wise Big Ideas Math Geometry Chapter 11 Solution Key and solve all easy & complex questions of Circumference, Area, and Volume with ease. Also, you can excel in math concepts by practicing the** BIM Geometry Textbook Answer Key**. Ace up your preparation with BIM Geometry Ch 11 Circumference, Area, and Volume Answer Key and check your knowledge accordingly for better scores and math skills.

### Circumference, Area, and Volume Maintaining Mathematical Proficiency

Find the surface area of the prism.

Question 1.

Answer:

The surface area of the prism = 158.

Explanation:

In the above-given question,

given that,

l = 5 ft, w = 8 ft, and h = 3 ft.

where l = length, w = width, and h = height.

The surface area of the rectanguler prism = 2(lw + lh + wh).

surface area = 2(5×8 + 5×3 + 8×3).

surface area = 2(40 + 15 + 24).

surface area = 2(79).

surface area = 158.

Question 2.

Answer:

The surface area of the triangular prism = 68m.

Explanation:

In the above-given question,

given that,

l = 10 m, p = 4 m, and h = 10.

the surface area of the triangular prism = 2B + ph.

b = base, p = perimeter, and h = height.

surface area = 2(6 + 8) + 4(10).

surface area = 2(14) + 40.

surface area = 28 + 40.

surface area = 68m.

Question 3.

Answer:

The surface area of the triangular prism = 42m.

Explanation:

In the above-given question,

given that,

w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.

the surface area of the triangular prism = 2B + ph.

b = base, p = perimeter, and h = height.

surface area = 2(6 + 5) + 4(5).

surface area = 2(11) + 20.

surface area = 22 + 20.

surface area = 42cm.

Find the missing dimension.

Question 4.

A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

Answer:

The length of the rectangle = 9 in.

Explanation:

In the above-given question,

given that,

A rectangle has a perimeter of 28 inches and a width of 5 inches.

length of the rectangle = p/2 – w.

length = 28/2 – 5.

where perimeter = 28 in, and w = 5 in.

length = 14 – 5.

length = 9.

so the length of the rectangle = 9 in.

Question 5.

A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

Answer:

The base of the triangle = 2 cm.

Explanation:

In the above-given question,

given that,

A triangle has an area of 12 sq cm and a height of 12 cm.

The base of the triangle = 2(A)/h.

base = 2(12)/12.

base = 24/12.

base = 2cm.

so the base of the triangle = 2 cm.

Question 6.

A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

Answer:

The length of the rectangle = 12 ft.

Explanation:

In the above-given question,

given that,

A rectangle has an area of 84 sq ft and a width of 7 feet.

area of the rectangle = l x b.

84 = l x 7.

l = 84/7.

l = 12.

so the length of the rectangle = 12 ft.

Question 7.

**ABSTRACT REASONING**

Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

Answer:

The surface area of a prism = 2(lw + wh + lh).

Explanation:

In the above-given question,

given that,

length = l, width = w, and height = x inches.

the surface area of the prism = 2(lw + wh + lh).

the solid figure does the prism represent the rectangular prism.

### Circumference, Area, and Volume Monitoring Progress

Draw a net of the three-dimensional figure. Label the dimensions.

Question 1.

Answer:

The surface area of the prism = 64 cm.

Explanation:

In the above-given question,

given that,

l = 2 ft, w = 4 ft, and h = 4 ft.

where l = length, w = width, and h = height.

The surface area of the rectanguler prism = 2(lw + lh + wh).

surface area = 2(2×4 + 4×4 + 4×2).

surface area = 2(8 + 16 + 8).

surface area = 2(32).

surface area = 64.

Question 2.

Answer:

The surface area of the prism = 392 m.

Explanation:

In the above-given question,

given that,

l =8 m, w = 12 m, and h = 5 m.

where l = length, w = width, and h = height.

The surface area of the rectanguler prism = 2(lw + lh + wh).

surface area = 2(8×12 + 12×5 + 5×8).

surface area = 2(96 + 60 + 40).

surface area = 2(196).

surface area = 392.

Question 3.

Answer:

The surface area of the triangular prism = 170 sq. in.

Explanation:

In the above-given question,

given that,

B = 10 in, p = 15 in, and h = 15 in, l = 10 in.

the surface area of the triangular prism = 2B + ph.

b = base, p = perimeter, and h = height.

surface area = 2(10) + 15(10).

surface area = 2(10) + 150.

surface area = 20 + 150.

surface area = 170 sq. in.

11.1 Circumference and Arc Length

**Exploration 1**

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

a. entire circle

Answer:

The Formula for the Arc Length is 2r(θ/360)

r = 4

θ = 260 degrees

Arc length = 2(4)(360/360)

= 8(1)

= 8

Therefore the Arc length is 8.

b. one-fourth of a circle

Answer:

The Formula for the Arc Length is 2r(θ/360)

r = 4

θ = 90 degrees.

Arc length = 2(4)(90/360)

= 8(90/360)

= 2

Therefore the Arc length is 2.

c. one-third of a circle

Answer:

The Formula for the Arc Length is 2r(θ/360)

r = 5

θ = 100 degrees.

Arc length = 2(5)(100/360)

= 10(100/360)

= 2.7

Therefore the Arc length is 2.7

d. five-eights of a circle

Answer:

The Formula for the Arc Length is 2r(θ /360)

r = 3

θ = 225 degrees.

Arc length = 2(3)(225/360)

= 6(225/360)

= 3.75

Therefore the Arc length is 3.75 cm

**Exploration 2**

Using Arc Length

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.

Answer:

C = πd

C = 25π or 78.54 inch

Half of revolution = 1/2 (78.54) = 39.27 inch

No, the time has gone past the box by 3.27 inch

Communicate Your Answer

Question 3.

How can you find the length of a circular arc?

Answer:

The length of a circular arc = 2

**LOOKING FOR REGULARITY IN REPEATED REASONING**

To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.

Question 4.

A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?

Answer:

The diameter of the motorcycle is 24 inches.

One revelation = 360 degrees.

Three-fourths of one revelation = 270 degrees.

The motorcycles travel = 2r(θ/360)

Radius r = d/2 = 24/2 = 12

θ = 270 degrees.

= 2(12)(270/360)

= 24(270/360)

= 18

Therefore the motorcycle travels 18 cm.

### Lesson 11.1 Circumference and Arc Length

**Monitoring Progress**

Question 1.

Find the circumference of a circle with a diameter of 5 inches.

Answer:

Circumference C = πd

C = 3.14x 5 = 15.7 in

Question 2.

Find the diameter of a circle with a circumference of 17 feet.

Answer:

Diameter d = C/π

d = 17/π = 5.41 ft

Find the indicated measure.

Question 3.

arc length of \(\widehat{P Q}\)

Answer:

arc length of \(\widehat{P Q}\) is 5.887

Explanation:

\(\widehat{P Q}\) = \(\frac { 75 }{ 360 } \) . π(9)

= 5.887

Question 4.

circumference of ⊙N

Answer:

arc length of LM/C = LM/360

61.26/C = 270/360

C = 81.68

Question 5.

radius of ⊙G

Answer:

arc length of EF = \(\frac { 60 }{ 360 } \) • 2πr

10.5 = \(\frac { 1 }{ 6 } \) • 2πr

r = 10.02

Question 6.

A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

Answer:

The car tire have to make 69 revolutions to travel 500 ft.

Explanation:

Circumference C = 2πr = πd

C = 28π

Distance travelled = number of revolutions x C

500 x 12 = number of revolutions x 28π

number of revolutions = 68.2

Question 7.

In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Answer:

Given that,

The arc radius for a runner on the blue path is 44.02 meters.

The diameter of the track is 2r = 2(44.02) = 88.04

The circumference of the track is πd = π(88.04) = 276.44

The runner travels around the track 276.44 cm.

Question 8.

Convert 15° to radians.

Answer:

15° = 15 . \(\frac { π radians }{ 180° } \) = \(\frac { π }{ 12 } \) radians

Question 9.

Convert \(\frac{4 \pi}{3}\) radians to degrees.

Answer:

\(\frac{4 \pi}{3}\) radians = \(\frac{4 \pi}{3}\) radians . \(\frac { 180° }{ π radians } \) = 240 degrees

### Exercise 11.1 Circumference and Arc Length

Vocabulary and Core Concept Check

Question 1.

**COMPLETE THE SENTENCE**

The circumference of a circle with diameter d is C = _______ .

Answer:

Question 2.

**WRITING**

Describe the difference between an arc measure and an arc length.

Answer:

An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 3.

circumference of a circle with a radius of 6 inches

Answer:

Question 4.

diameter of a circle with a circumference of 63 feet

Answer:

C = 63 ft

πd = 63

d = 20.05

Question 5.

radius of a circle with a circumference of 28π

Answer:

Question 6.

exact circumference of a circle with a diameter of 5 inches

Answer:

C = πd

C = 5π = 15.707

Question 7.

arc length of \(\widehat{A B}\)

Answer:

Question 8.

m\(\widehat{D E}\)

Answer:

\(\frac { arc length of DE }{ 2πr } \) = \(\frac { DE }{ 360 } \)

\(\frac { 8.73 }{ 2π(10) } \) = \(\frac { DE }{ 360 } \)

DE = 50.01°

Question 9.

circumference of ⊙C

Answer:

Question 10.

radius of ⊙R

Answer:

\(\frac { arc length of LM }{ 2πr } \) = \(\frac { LM }{ 360 } \)

\(\frac { 38.95 }{ 2πr } \) = \(\frac { 260 }{ 360 } \)

r = 8.583

Question 11.

**ERROR ANALYSIS**

Describe and correct the error in finding the circumference of ⊙C.

Answer:

Question 12.

**ERROR ANALYSIS**

Describe and correct the error in finding the length of \(\widehat{G H}\).

Answer:

\(\frac { arc length of GH }{ 2πr } \) = \(\frac { m GH }{ 360 } \)

\(\widehat{G H}\). = \(\frac { 5 }{ 24 } \) . 2π(10)

= 13.08

Question 13.

**PROBLEM SOLVING**

A measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?

Answer:

Question 14.

**PROBLEM SOLVING**

You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?

Answer:

Circumference of the front wheel = 2π(32.5)

= 65π cm

Distance covered = 40 m = 40 x 100 = 4000 cm

Number of revolutions = \(\frac { 4000 }{ 65π } \) = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

Question 15.

Answer:

Question 16.

Answer:

Two horizontal edges are 2 . 3 = 6

Circumference of circle = 2π(3) = 6π

The perimeter of the shaded region = 6 + 6π

Question 17.

Answer:

Question 18.

Answer:

Length of the shape excluding three arcs = 5 + 5 + 5 = 15

Radius of the arc = 2.5

Opposite sides are equal so the length from midpoint of one circle to another is 5.

Arc length = 120/360 × 2π(2.5)

= 1/3 × 5 × 3.14

= 5 × 3.14/3

= 5.23

perimeter of the shaded region = length of shape excluding three arcs + three times arc length

= 15 + 3(5.23)

= 15 + 15.69

= 30.69

In Exercises 19 – 22, convert the angle measure.

Question 19.

Convert 70° to radians.

Answer:

Question 20.

Convert 300° to radians.

Answer:

300 • (\(\frac { π }{ 180 } \)) = \(\frac { 5π }{ 3 } \) radian

Question 21.

Convert \(\frac{11 \pi}{12}\) radians to degrees.

Answer:

Question 22.

Convert \(\frac{\pi}{8}\) radian to degrees.

Answer:

\(\frac { π }{ 8 } \) • \(\frac { 180 }{ π } \)) = 22.5°

Question 23.

**PROBLEM SOLVING**

The London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?

Answer:

Question 24.

**PROBLEM SOLVING**

You are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.

(A) 7

(B) 8

(C) 9

(D) 10

Answer:

C = 38 ft

2πr = 38

r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 25.

x^{2} + y^{2} = 16

Answer:

Question 26.

(x + 2)^{2} + (y – 3)^{2} = 9

Answer:

The radius of circle (x + 2)² + (y – 3)² = 9 is 3

C = 2πr = 2π(3) = 6π

The circumference of the circle is 6π units.

Question 27.

**USING STRUCTURE**

A semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.

Answer:

Question 28.

**REASONING**

\(\widehat{E F}\) is an arc on a circle with radius r. Let x° be the measure of \(\widehat{E F}\). Describe the effect on the length of \(\widehat{E F}\) if you (a) double the radius of the circle, and (b) double the measure of \(\widehat{E F}\).

Answer:

Given x° is the measure of \(\widehat{E F}\)

arc length of a circle = x/360 × 2πr

arc length of \(\widehat{E F}\) = x/360 × 2πr

a. double the radius of the circle

x/360 × 2π(2r) = 2x/360 × 2πr

x/360 × 2π(2r) = 2arc length of \(\widehat{E F}\)

(b) double the measure of \(\widehat{E F}\)

2x/360 × 2πr = 2 × x/360 × 2πr

2x/360 × 2πr = 2arc length \(\widehat{E F}\)

Question 29.

**MAKING AN ARGUMENT**

Your friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.

Answer:

Question 30.

**PROBLEM SOLVING**

Over 2000 years ago, the Greek scholar Eratosthenes estimated Earth’s circumference by assuming that the Sun’s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun’s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth’s circumference. Then estimate Earth’s circumference.

Answer:

∠1 and ∠2 are the alternate angles

m∠2 = m∠1 = 7.2°

2000 ÷ 575 = 3.47

Circumference = 2 × 3.14 × 3.47 = 21.8 + 7.2 = 29 (approx)

So, the estimated earths circumference is 29 miles.

Question 31.

**ANALYZING RELATIONSHIPS**

In ⊙C the ratio of the length of \(\widehat{P Q}\) to the length of \(\widehat{R S}\) is 2 to 1. What is the ratio of m∠PCQ to m∠RCS?

(A) 4 to 1

(B) 2 to 1

(C) 1 to 4

(D) 1 to 2

Answer:

Question 32.

**ANALYZING RELATIONSHIPS**

A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r_{1} of ⊙C to the radius r_{2} of ⊙P? Explain your reasoning.

Answer:

Given,

A 45° arc in ⊙C and a 30° arc in ⊙P have the same length.

r1/r2 = 45/30 = 3/2

So, the ratio of the radius r_{1} of ⊙C to the radius r_{2} of ⊙P is 3/2.

Question 33.

**PROBLEM SOLVING**

How many revolutions does the smaller gear complete during a single revolution of the larger gear?

Answer:

Question 34.

**USING STRUCTURE**

Find the circumference of each circle.

a. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long

Answer:

c² = a² + b²

c² = 12²+ 16²= 400

c = 20 in

circumference c = dπ

= 20π = 62.83 in

b. a circle circumscribed about a square with a side length of 6 centimeters

Answer:

d² = 6²+ 6² = 72

d = 8.49 cm

C = dπ

C = 8.49π

C = 26.67 cm

c. a circle inscribed in an equilateral triangle with a side length of 9 inches

Answer:

r = \(\frac { a√3 }{ 3 } \)

r = \(\frac { 9√3 }{ 3 } \)

r = 3√3 = 5.2

C = 2πr

C = 2π (5.2) = 32.67 in

Question 35.

**REWRITING A FORMULA**

Write a formula in terms of the measure θ (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of \(\frac{3 \pi}{4}\) radians.

Answer:

Question 36.

**HOW DO YOU SEE IT?**

Compare the circumference of ⊙P to the length of \(\widehat{D E}\). Explain your reasoning.

Answer:

The radius of C is 2r.

The arc length DE is subtended by a straight angle and so can be evaluated as

180/360 × 2π × 2r = 2πr

Comparing the circumference of P we can see that it is equal to the arc length of DE.

Question 37.

**MAKING AN ARGUMENT**

In the diagram. the measure of the red shaded angle is 30°. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.

Answer:

Question 38.

**MODELING WITH MATHEMATICS**

What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.

a. 1 : 30 P.M.

Answer:

3π/4

b. 3:15 P.M.

Answer:

π/24

Question 39.

**MATHEMATICAL CONNECTIONS**

The sum of the circumferences of circles A, B, and C is 63π. Find AC.

Answer:

Question 40.

**THOUGHT PROVOKING**

Is π a rational number? Compare the rational number \(\frac{355}{113}\) to π. Find a different rational number that is even closer π.

Answer:

π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimal places of π, therefore 3.1415926 x x = a.

Question 41.

**PROOF**

The circles in the diagram are concentric and \(\overrightarrow{F G}\) ≅ \(\overrightarrow{G H}\) Prove that \(\widehat{J K}\) and \(\widehat{N G}\) have the same length.

Answer:

Question 42.

**REPEATED REASONING**

\(\overline{A B}\) is divided into four congruent segments, and semicircles with radius r are drawn.

a. What is the sum of the four arc lengths?

Answer:

360/2 = 180 degrees

Then the arc length of 1 semicircle is

180/360 × 2πr

1/2 × 2πr = πr

Therefore the arc length of 4 semicircles is 4 × πr = 4πr

The sum of the four arc lengths is 4πr units.

b. What would the sum of the arc lengths be if \(\overline{A B}\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.

Answer:

360/2 = 180 degrees

Then the arc length of 1 semicircle is

180/360 × 2π(r/2)

1/2 × πr = πr/2

Therefore the arc length of 8 semicircles will be

8 × πr/2 = 4πr

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 43.

X(2, 4), Y(8, – 1), Z(2, – 1)

Answer:

Question 44.

L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

Answer:

LP = √(-3 + 3)² + (-5 – 1)² = 6

PN = √(4 + 3)² + (-5 + 5)² = 7

MN = √(4 – 4)² + (-5 – 1)² = 6

LM = √(4 + 3)² + (1 – 1)²= 7

Area = 6 x 7 = 42 units

### 11.2 Areas of Circles and Sectors

**Exploration 1**

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

a. entire circle

Answer:

The Formula for the area of the sector is πr²(θ/360)

r = 4

θ = 360 degrees.

Arc length = π(4)²(360/360)

= π(16)(360/360)

= 16π

= 50.24 cm

Therefore the area of the shaded circle or sector of a circle is 50.24 sq. cm.

b. one – fourth of a circle

Answer:

The Formula for the area of the sector is πr²(θ/360)

r = 5

θ = 90 degrees.

Arc length = π(5)²(90/360)

= π(25)(90/360)

= 6.25π

= 19.625 cm

Therefore the area of the shaded circle or sector of a circle is 19.625 sq. cm.

c. seven – eighths of a circle.

Answer:

The Formula for the area of the sector is πr²(θ/360)

r = 2

θ = 315 degrees.

Arc length = π(4)²(360/360)

= π(4)(315/360)

= 3.5π

= 10.99 cm

Therefore the area of the shaded circle or sector of a circle is 10.99 sq. cm.

d. two – thirds of a circle

Answer:

The Formula for the area of the sector is πr²(θ/360)

r = 8

θ = 230 degrees.

Arc length = π(8)²(230/360)

= π(64)(230/360)

= 40.88π

= 128.36 cm

Therefore the area of the shaded circle or sector of a circle is 128.36 sq. cm.

Exploration 2

Finding the Area of a Circular Sector

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.

**REASONING ABSTRACTLY**

To be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.

Answer:

Communicate Your Answer

Question 3.

How can you find the area of a sector of a circle?

Answer:

The formula for sector area is simple, just multiply the central angle by the radius squared, and divide by 2

Area of a sector = θ/360 × πr²

Question 4.

In Exploration 2, find the area of the sector that is irrigated in 2 hours.

Answer:

### Lesson 11.2 Areas of Circles and Sectors

**Monitoring progress**

Question 1.

Find the area of a circle with a radius of 4.5 meters.

Answer:

Circle area = πr²

A = π(4.5)² = 20.25π

Question 2.

Find the radius of a circle with an area of 176.7 square feet.

Answer:

Circle area = πr²

176.7 = πr²

r² = 56.24

r = 7.499

Question 3.

About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

Answer:

The population density is about 4615.49 people per square mile.

Explanation:

A = πr² = π • 2² = 4π

Population density = \(\frac { number of people }{ area of land } \)

= \(\frac { 58000 }{ 4π } \) = 4615.49

Question 4.

A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

Answer:

The number of people who live in the region are 28274.

Explanation:

A = πr² = π • 3² = 9π

Population density = \(\frac { number of people }{ area of land } \)

Number of people = 1000 x 9π = 28274

Find the indicated measure

Question 5.

area of red sector

Answer:

The area of red sector = 205.25

Explanation:

m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240°

Area of red sector = \(\frac { FE }{ 360° } \) • πr²

= \(\frac { 120 }{ 360° } \) • π(14²)

= 205.25

Question 6.

area of blue sector

Answer:

Area of blue sector = 410.5

Explanation:

Area of blue sector = \(\frac { FGE }{ 360° } \) • πr²

= \(\frac { 240 }{ 360° } \) • π(14²) = 410.5

Question 7.

Find the area of ⊙H.

Answer:

Area of ⊙H = 907.92 sq cm

Explanation:

Area of sector FHG =\(\frac { FG }{ 360° } \) • Area of ⊙H

214.37 = \(\frac { 85 }{ 360° } \) • Area of ⊙H

Area of ⊙H = 907.92 sq cm

Question 8.

Find the area of the figure.

Answer:

Area of triangle = \(\frac { 1 }{ 2 } \) • 7 • 7

= 24.5 sq m

Area of semi circle = πr²/2

= π(3.5)²/2

= 19.242255

Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9.

If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

Answer:

Yes, we can find the intercepted arc when we the area and the radius of the sector of the circle.

Because the intercepted arc is the arc inside the inscribed angles and whose endpoints are on the angle.

### Exercise 11.2 Areas of Circles and Sectors

Vocabulary and Core Concept Check

Question 1.

**VOCABULARY**

A(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.

Answer:

Question 2.

**WRITING**

The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Answer:

Yes

Explanation:

Area of sector with arc measure x and radius r is s = π/180(xr)

If x becomes doube, then s1 = π/180(2xr) = 2s

This means that if the arc measure doubles, area of the sector also doubles.

Monitoring Progress and Modeling with Mathematics

In Exercise 3 – 10, find the indicated measure,

Question 3.

area of ⊙C

Answer:

Question 4.

area of ⊙C

Answer:

Area A = πr²

A = π(10)² = 100π sq in

Question 5.

area of a circle with a radius of 5 inches

Answer:

Question 6.

area of a circle with a diameter of 16 feet

Answer:

d = 2r

Circle area = πr² = (π/4)d²

= (π/4)16² = 64π

Question 7.

radius of a circle with an area of 89 square feet

Answer:

Question 8.

radius of a circle with an area of 380 square inches

Answer:

A = πr²

380 = πr²

r = 10.99

Question 9.

diameter of a circle with an area of 12.6 square inches

Answer:

Question 10.

diameter of a circle with an area of 676π square centimeters

Answer:

Area A = 676π square centimeters

(π/4)d² = 676π

d² = 2704

d = 52

In Exercises 11 – 14, find the indicated measure.

Question 11.

About 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.

Answer:

Question 12.

About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

Answer:

The population density is about 5747 people per square mile.

Explanation:

Area of region = π(6)² = 36π

Population density = \(\frac { Number of people }{ area of land } \)

= \(\frac { 650,000 }{ 36π } \) = 5747.2

Question 13.

A region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.

Answer:

Question 14.

About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

Answer:

The radius of the region is 7

Explanation:

Population density = \(\frac { Number of people }{ area of land } \)

513 = \(\frac { 79,000 }{ πr² } \)

πr² = 153.99

r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Question 15.

Answer:

Question 16.

Answer:

Area of sector = \(\frac { 104° }{ 360° } \) • π(14)²

= 177.88

Area of red region is 177.88 sq cm

Area of blue region = \(\frac { 256° }{ 360° } \) • π(14)²

= 437.86 sq cm

Question 17.

Answer:

Question 18.

Answer:

Area of red region is 10.471 sq ft

Area of the blue region is 39.79 sq ft

Explanation:

Area of sector = \(\frac { 75° }{ 360° } \) • π(4)²

= 10.471

Area of red region is 10.471 sq ft

Area of blue region = \(\frac { 285° }{ 360° } \) • π(4)²

= 39.79 sq ft

Question 19.

**ERROR ANALYSIS**

Describe and correct the error in finding. the area of the circle.

Answer:

Question 20.

**ERROR ANALYSIS**

Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet.

Answer:

Area of ⊙Z is 255 square feet

πr² = 255

r = 9

Area of sector XZY = \(\frac { 115 }{ 360 } \) • 255

n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Question 21.

area of ⊙M

Answer:

Question 22.

radius of ⊙M

Answer:

radius of ⊙M = 3.98

Explanation:

Area of region = \(\frac { 89 }{ 360 } \) . Area of ⊙M

12.36 = \(\frac { 89 }{ 360 } \) . Area of ⊙M

Area of ⊙M = 49.99

πr² = 49.99

r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Question 23.

Answer:

Question 24.

Answer:

The area of the shaded region is 85.840 sq in.

Explanation:

Area of square = 20² = 400

Diameter of one circle = 10

radius of one circle = 5 in

Area of one circle = π(5)² = 78.53

Areas of four circle = 314.159

Area of shaded region = 400 – 314.159 = 85.840

Question 25.

Answer:

Question 26.

Answer:

The area of shaded region is 301.59

Explanation:

The radius of smaller circle is 8 cm

The radius of bigger circle is 16 cm

Area of smaller semicircle = \(\frac { 1 }{ 2 } \)(π(8)²) = 100.53

Area of lager semicircle = \(\frac { 1 }{ 2 } \)(π(16)²) = 402.123

Area of shaded region = 402.123 – 100.53 = 301.59

Question 27.

Answer:

Question 28.

Answer:

Area of shaded region = 7.63

Explanation:

c² = 3² + 4² = 25

c = 5

Radius = 2.5

Circle area = π(2.5)² = 19.63

Area of triangle = (3 x 4)/2 = 6

Area of shaded region = 19.63 – 12 = 7.63

Question 29.

**PROBLEM SOLVING**

The diagram shows the shape of a putting green at a miniature golf course. One part of the green is a sector of a circle. Find the area of the putting green.

Answer:

Question 30.

**MAKING AN ARGUMENT**

Your friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.

Answer:

The friend is not correct. doubling the radius quadruples the area.

Explanation:

Area of circle with radius r = πr²

Area of circle with radius 2r = π(2r)² = 4πr²

Question 31.

**MODELING WITH MATHEMATICS**

The diagram shows the area of a lawn covered by a water sprinkler.

a. What is the area of the lawn that is covered by the sprinkler?

b. The water pressure is weakened so that the radius is 12 feet. What is the area of the lawn that will be covered?

Answer:

Question 32.

**MODELING WITH MATHEMATICS**

The diagram shows a projected beam of light from a lighthouse.

a. What is the area of water that can be covered by the light from the lighthouse?

Answer:

Area = \(\frac { 115 }{ 360 } \) x π(18)²

= 325.15 sq mi

b. What is the area of land that can be covered by the light from the lighthouse?

Answer:

Area = \(\frac { 245 }{ 360 } \) x π(18)²

= 692.72 sq mi

Question 33.

**ANALYZING RELATIONSHIPS**

Look back at the Perimeters of Similar Polygons Theorem (Theorem 8.1) and the Areas of Similar PoIyons Theorem (Theorem 8.2) in Section 8.1. How would you rewrite these theorems to apply to circles? Explain your reasoning.

Answer:

Question 34.

**ANALYZING RELATIONSHIPS**

A square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.

Answer:

We start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is πr²

It can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r.

Next, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r√2. outer circle radius = r√2

Area of outer circle 2πr²

The ratio of the area of larger circle to the smaller circle = 2.

Question 35.

**CONSTRUCTION**

The table shows how students get to school.

a. Explain why a circle graph is appropriate for the data.

b. You will represent each method by a sector of a circle graph. Find the central angle to use for each sector. Then construct the graph using a radius of 2 inches.

c. Find the area of each sector in your graph.

Answer:

Question 36.

**HOW DO YOU SEE IT?**

The outermost edges of the pattern shown form a square. If you know the dimensions of the other square, is it possible to compute the total colored area? Explain.

Answer:

It is divided into two parts:

First is square

Second is 1/4 of a circle

So we can compute the colored area by calculating the areas of squares plus 1/4 circles

So, it is possible.

Question 37.

**ABSTRACT REASONING**

A circular pizza with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs $12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain.

a. You want to spend as little money as possible.

b. You want to have three pizzas. each with a different topping, and spend as little money as possible.

C. You want to have as much of the thick outer crust as possible.

Answer:

Question 38.

**THOUGHT PROVOKING**

You know that the area of a circle is πr^{2}. Find the formula for the area of an ellipse, shown below.

Answer:

Ellipse area = πab

Question 39.

**MULTIPLE REPRESENTATIONS**

Consider a circle with a radius of 3 inches.

a. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth.

b. Graph the data in the table.

c. Is the relationship between x and y linear? Explain.

d. If parts (a) – (c) were repeated using a circle with a radius of 5 inches, would the areas in the table change? Would your answer to part (c) change? Explain your reasoning.

Answer:

Question 40.

**CRITICAL THINKING**

Find the area between the three congruent tangent circles. The radius of each circle is 6 inches.

Answer:

Number of circles = 3

radius = 6 in.

Using Pythagoras theorem,

12² = h² + 6²

h² = 144 – 36

h² = 108

h = 6√3

Area of equilateral triangle

A = 1/2 × bh

h = 6√3

b = 2r = 2 × 6 = 12

A = 1/2 × 12 × 6√3

A1 = 36√3

Since the radii formed an equilateral triangle, then the central angle will be 60.

A = θ/360 × πr²

A = 60/360 × π × 36

A = 60/10 × π

A = 6π

For three circles the area is

A2 = 3 × 6π

A2 = 18π

A = A1 – A2

A = 36√3 – 18π

A = 36 × 1.7321 – 18 × 3.14

A = 5.8356 sq. in

Question 41.

**PROOF**

Semicircles with diameters equal to three sides of a right triangle are drawn, as shown. Prove that the sum of the areas of the two shaded crescents equals the area of the triangle.

Answer:

Maintaining Mathematical proficiency

Find the area of the figure.

Question 42.

Answer:

Area = \(\frac { 1 }{ 2 } \)(base x height)

Area = \(\frac { 1 }{ 2 } \)(18 x 6) = 54 sq in

Question 43.

Answer:

Question 44.

Answer:

Area = \(\frac { 1 }{ 2 } \)(base x height)

A = \(\frac { 1 }{ 2 } \)(13 x 9) = 58.5 sq in

Question 45.

Answer:

### 11.3 Areas of Polygons

**Exploration 1**

Finding the Area of a Regular Polygon

Work with a partner: Use dynamic geometry software to construct each regular polygon with side lengths of 4, as shown. Find the apothem and use it to find the area of the polygon. Describe the steps that you used.

a.

Answer:

Given that the polygon is in the shape of a triangle.

The side length is 4.

Height = 3.5

The formula for the area of the triangle is ½ x b x h

= 1/2 x 4 x 3.5

= 14/2

= 7 square cm

Therefore the area of the triangle is 7 square cm.

b.

Answer:

Given that,

The formula for the area of the pentagon if only the side length is known is A = 1/4 x square root(5 + 2(square root of 5) x a²

The side length is a = 4

A = 1/4 x square root of 5(5 + 2 square root(5) x (4)²

= 1/4 x square root of 5 + 4.47 x 16

= 1/4 x square root of (9.47) x 16

= 1/4 x 3.077 x 16

= 1/4 x 49.232

= 12.308 square cm

Therefore the area of the pentagon is 12.308 square cm.

c.

Answer:

Given that the polygon is in the shape of a hexagon.

The side length is 4.

The formula for the area of the hexagon is 3 square root(3) x s²/2

= 3 square root(3) x 16/2

= 41.56 square cm

Therefore the area of the hexagon is 41.56 square cm.

d.

Answer:

Given that the polygon is in the shape of an octagon.

The side length is 4.

The formula for the area of the octagon is A = 8/2 x a x h

a = length

h = height

= 8/2 x 4 x 3.5

= 8/2 x 14

= 56 square cm

Therefore the area of the octagon is 56 square cm.

**Exploration 2**

Writing a Formula for Area

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon.

**REASONING ABSTRACTLY**

To be proficient in math, you need to know and flexibly use different properties of operations and objects.

Answer:

Communicate Your Answer

Question 3.

How can you find the area of a regular polygon?

Answer:

You can find the area of the regular pentagon using the formulas.

They are,

The formula for the regular pentagon if only the side is known is A = 1/4 x square root of 5(5 + 2 square root(5) x (a)².

The formula for the area of the regular pentagon is 1/2 x p x a.

Where a = apothem

P = perimeter

Question 4.

Regular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE.

Answer:

Given that,

The side length of the regular pentagon ABCDE is 6 meters.

The apothem of the regular pentagon is 4.13 meters.

The formula for the area of the regular pentagon is 1/2 x p x a.

Where

a = apothem

P = perimeter

The formula for the perimeter of a regular pentagon is = 5a

= 5(6) = 30

= 1/2 x 30 x 4.13

= 1/2 x 123.9

= 61.95 square cm

### Lesson 11.3 Areas of Polygons

**Monitoring Progress**

Question 1.

Find the area of a rhombus with diagonals d_{1} = 4 feet and d_{2} = 5 feet.

Answer:

Area of rhombus = \(\frac { 1 }{ 4 } \)(d₁d₂)

= \(\frac { 1 }{ 4 } \)(4 x 5) = 5 sq ft

Question 2.

Find the area of a kite with diagonals d_{1} = 12 inches and d_{1} = 9 inches.

Answer:

Area of kite = \(\frac { 1 }{ 4 } \)(d₁d₂)

= \(\frac { 1 }{ 4 } \)(12 x 9) = 27 sq in

In the diagram. WXYZ is a square inscribed in ⊙P.

Question 3.

Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:

P is the center, PY or PX is the radius, PQ is apothem, ∠XPY is the central angle.

Question 4.

Find m∠XPY, m∠XPQ, and m∠PXQ.

Answer:

m∠XPY = \(\frac { 360 }{ 4 } \) = 90

m∠XPQ = 90/2 = 45

m∠PXQ = 180 – (90 + 45) = 45

Find the area of regular polygon

Question 5.

Answer:

c = √(8² + 6.5²) = 10.3

a = 20.61

Area = 0.25(√5(5+2√5) a²

Area = 0.25(√5(5+2√5) 20.61² = 730.8

Question 6.

Answer:

Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)

= \(\frac { 5(7²) }{ 2 } \)√(5+2√5)

Area = 55377

### Exercise 11.3 Areas of Polygons

Vocabulary and Core Concept Check

Question 1.

**WRITING**

Explain how to find the measure of a central angle of a regular polygon.

Answer:

Question 2.

**DIFFERENT WORDS, SAME QUESTION**

Which is different? Find “both” answers.

Find the radius of ⊙F. Answer:

Answer:

EF = radius = 6.8

Find the apothem of polygon ABCDE.

Answer:

GF = apothem = 5.5

Find AF.

Answer:

AF = √4² + 5.5²

AF = 6.8

Find the radius of polygon ABCDE.

Answer:

AF = radius = 6.8

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the area of the kite or rhombus.

Question 3.

Answer:

Question 4.

Answer:

d₁ = 6 + 6 = 12

d₂ = 2 + 10 = 12

area A = \(\frac { 1 }{ 4 } \)(d₁d₂)

= \(\frac { 1 }{ 4 } \)(12 x 12)

= 36

Question 5.

Answer:

Question 6.

Answer:

area A = \(\frac { 1 }{ 4 } \)(d₁d₂)

A = \(\frac { 1 }{ 4 } \)(5 x 6) = 7.5

In Exercises 7 – 10, use the diagram

Question 7.

Identify the center of polygon JKLMN?

Answer:

Question 8.

Identify a central angle of polygon JKLMN.

Answer:

∠NPM is the central angle of polygon JKLMN

Question 9.

What is the radius of polygon JKLMN?

Answer:

Question 10.

What is the apothem of polygon JKLMN?

Answer:

QP is the apothem of polygon JKLMN

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.

Question 11.

10 sides

Answer:

Question 12.

18 sides

Answer:

The measure of central angle = \(\frac { 360 }{ 18 } \) = 20

Question 13.

24 sides

Answer:

Question 14.

7 sides

Answer:

The measure of central angle = \(\frac { 360 }{ 7 } \) = 51.42

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.

Question 15.

m∠GJH

Answer:

Question 16.

m∠GJK

Answer:

m∠GJK = m∠GJH/2

m∠GJK = 22.5

Question 17.

m∠KGJ

Answer:

Question 18.

m∠EJH

Answer:

m∠EJH = 3(45) = 135

In Exercises 19 – 24, find the area of the regular polygon.

Question 19.

Answer:

Question 20.

Answer:

Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)

Area = \(\frac { 5(6.84)² }{ 2 } \)√(5+2√5)

A = 359.9784

Question 21.

Answer:

Question 22.

Answer:

Area = \(\frac { 3√3 a²}{ 2 } \)

A = \(\frac { 3√3 (7)²}{ 2 } \)

A = 127.30

Question 23.

an octagon with a radius of 11 units

Answer:

Question 24.

a pentagon with an apothem of 5 units

Answer:

A = 90.75

Explanation:

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360/5 = 72

After that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function.

tan y = p/a

tan 36 = p/5

p = 3.63

Since p is just half of the length of the side, we have to multiply it by 2

2 . p = 2 . 3.63 = 7.26 = s

Area = \(\frac { a . s. n }{ 2 } \)

A = \(\frac { 15 x 7.26 x 5 }{ 2 } \)

A = 90.75

Question 25.

**ERROR ANALYSIS**

Describe and correct the error in finding the area of the kite.

Answer:

Question 26.

**ERROR ANALYSIS**

Describe and correct the error in finding. the area of the regular hexagon.

Answer:

s = √15² – 13² = 7.48

Area = \(\frac { 1 }{ 2 } \)(a . ns)

A = \(\frac { 1 }{ 2 } \)(13 x 6 x 7.48)

A = 291.72

In Exercises 27 – 30, find the area of the shaded region.

Question 27.

Answer:

Question 28.

Answer:

Area of the shaded region = 223.75

Explanation:

Square side = diagonal/√2

= 28/√2 = 19.79

Area of square = 19.79² = 392

Circle area = π(14)² = 615.75

Area of the shaded region = 615.75 – 392 = 223.75

Question 29.

Answer:

Question 30.

Answer:

Question 31.

**MODELING WITH MATHEMATICS**

Basaltic columns arc geological formations that result from rapidly cooling lava. Giant’s Causeway in Ireland contains many hexagonal basaltic columns. Suppose the top of one of the columns is in the shape of a regular hexagon with a radius of 8 inches. Find the area of the top of the column to the nearest square inch.

Answer:

Question 32.

**MODELING WITH MATHEMATICS**

A watch has a circular surface on a background that is a regular octagon. Find the area of the octagon. Then find the area of the silver border around the circular face.

Answer:

L = 1 cm

Area = 2L²/tan(22.5°)

= 2(1)²/tan(22.5)

Area = 4.828 sq. cm

silver border:

Silver border is a circular crown.

A = π(R² – r²)

where R = (1 + 0.2) = 1.2 cm

r = 1 cm

A = π(1.2² – 1²)

A = 3.14 × 0.44 = 1.38 sq. cm

**CRITICAL THINKING**

In Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning

Question 33.

The area of a regular n-gon of a fixed radius r increases as n increases.

Answer:

Question 34.

The apothem of a regular polygon is always less than the radius.

Answer:

true, the radius always reaches the end of the circle but the apothem never does

Question 35.

The radius of a regular polygon is always less than the side length.

Answer:

Question 36.

**REASONING**

Predict which figure has the greatest area and which has the least area. Explain your reasoning. Check by finding the area of each figure.

(A)

(B)

(C)

Answer:

(B) has the highest area and C has the lowest area.

Explanation:

(A) area = π(6.5)² = 132.73

(B) area = 139.25

(C) area = \(\frac { 1 }{ 2 } \)(18 x 15) = 135

Question 37.

**USING EQUATIONS**

Find the area of a regular

pentagon inscribed in a circle whose equation is given by (x – 4)^{2} + y + 2)^{2} = 25.

Answer:

Question 38.

**REASONING**

What happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.

Answer:

Area of a kite = \(\frac { 1 }{ 4 } \)(d₁d₂)

If you double the length of one diagonal, then d₁ = 2d₁

Area of kite = \(\frac { 1 }{ 2 } \)(d₁d₂)

If you double length of both diagonals

Area = \(\frac { 1 }{ 4 } \)(2d₁2d₂) = d₁d₂

If you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.

**MATHEMATICAL CONNECTIONS**

In Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.

Question 39.

The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.

Answer:

Question 40.

One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Answer:

The length of each diagonal is 9.89, 2.47.

Explanation:

One diagonal of a rhombus is four times the length of the other diagonal.

d₁ = 4d₂

Area = \(\frac { 1 }{ 4 } \)(d₁d₂)

98 = \(\frac { 1 }{ 4 } \)(d₁(4d₁))

d₁ = 9.89

d₂ = 2.47

Question 41.

**REASONING**

The perimeter of a regular nonagon. or 9-gon, is 18 inches. Is this enough information to find the area? If so, find the area and explain your reasoning. If not, explain why not.

Answer:

Question 42.

**MAKING AN ARGUMENT**

Your friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.

Answer:

No; A rhombus is not a regular polygon.

Question 43.

**PROOF**

Prove that the area of any quadrilateral with perpendicular diagonals is A = \(\frac{1}{2}\)d_{1}d_{2}, where d_{1} and d_{2} are the lengths of the diagonals.

Answer:

Question 44.

**HOW DO YOU SEE IT?**

Explain how to find the area of the regular hexagon by dividing the hexagon into equilateral triangles.

Answer:

Given that,

The hexagon has 6 sides.

The hexagon is divided into 6 equilateral triangles.

The area of the equilateral triangle is (square root of 3)/4 x a²

a = side length

= (square root of 3)/4 x (6)²

= (square root of 3)/4 x 36

= 15.588 square cm.

Question 45.

**REWRITING A FORMULA**

Rewrite the formula for the area of a rhombus for the special case of a square with side length s. Show that this is the same as the formula for the area of a square, A = s^{2}.

Answer:

Question 46.

**REWRITING A FORMULA**

Use the formula for the area of a regular polygon to show that the area of an equilateral triangle can be found by using the formula A = \(\frac{1}{4}\)s^{2}√3 where s is the side length.

Answer:

BE/ED = tan θ

Substitute θ as 60 degrees

BE/ED = tan 60°

BE = tan 60°(ED)

s/2 = (√3)a

a = s/2√3

A = 1/2 . a. (n . s)

A = 1/2 . s/2√3 . (3 . s)

A = 1/4 . √3 . s²

A = √3/4 s²

Question 47.

**CRITICAL THINKING**

The area of a regular pentagon is 72 square centimeters. Find the length of one side.

Answer:

Question 48.

**CRITICAL THINKING**

The area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.

Answer:

Let the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150/2 = 75. The adjacent to this angle is the length 2x/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is

140 = 1/2 (x tan75)(12 . 2x)

140 = 44.785 x²

x² = 3.126

x = 1.768

Question 49.

**USING STRUCTURE**

In the figure, an equilateral triangle lies inside a square inside a regular pentagon inside a regular hexagon. Find the approximate area of the entire shaded region to the nearest whole number.

Answer:

Question 50.

**THOUGHT PROVOKING**

The area of a regular n-gon is given by A = \(\frac{1}{2}\)aP. As n approaches infinity, what does the n-gon approach? What does P approach? What does a approach? What can you conclude from your three answers? Explain your reasoning.

Answer:

Question 51.

**COMPARING METHODS**

Find the area of regular pentagon ABCDE by using the formula A = \(\frac{1}{2}\)aP, or A = \(\frac{1}{2}\)a • ns. Then find the area by adding the areas of smaller polygons. Check that both methods yield the same area. Which method do you prefer? Explain your reasoning.

Answer:

Question 52.

**USING STRUCTURE**

Two regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.

Answer:

The radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360/n, therefore the apothem makes an angle of 180/n. Use sine and cosine to find the apothem and side length of the smaller polygon.

a_{small} = r sin\(\frac { 180 }{ n } \)

s_{small} = 2r cos\(\frac { 180 }{ n } \)

Use tangent to find the side length of the large polygon.

S_{large} = 2r tan\(\frac { 180 }{ n } \)

Use the formula to find the area of the smaller polygon.

A_{small} = 1/2 . a_{small} . n . s_{small}

A_{small} = 1/2 . r sin\(\frac { 180 }{ n } \) . n . 2r cos\(\frac { 180 }{ n } \)

A_{small} = nr² sin \(\frac { 180 }{ n } \) cos\(\frac { 180 }{ n } \)

Use the formula to find the area of the larger polygon.

A_{Large} = 1/2 . a_{large} . n . slarge

= nr² tan\(\frac { 180 }{ n } \)

The area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression.

A = A_{large} – A_{small}

A = nr² tan\(\frac { 180 }{ n } sin²[latex]\frac { 180 }{ n }

Maintaining Mathematical Proficiency

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the

figure has line symmetry. determine the number of lines of symmetry. It the figure has rotational

symmetry, describe any rotations that map the figure onto itself.

Question 53.

Answer:

Question 54.

Answer:

The figure has rotational symmetry.

Question 55.

Answer:

Question 56.

Answer:

The figure has one line symmetry.

### 11.4 Three-Dimensional Figures

**Exploration 1**

Analyzing a Property of Polyhedra

Work with a partner: The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.

Answer:

The tetrahedron has 4 vertices, 6 edges and 4 faces.

The cube has 8 vertices, 12 edges and 6 faces.

The octahedron has 6 vertices, 12 edges and 8 faces.

The dodecahedron has 230 vertices, 6 edges and 12 faces.

The icosahedron has 12 vertices, 30 edges and 20 faces.

Communicate Your Answer

Question 2.

What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.)

**CONSTRUCTING VIABLE ARGUMENTS**

To be proficient in math, you need to reason inductively about data.

Answer:

The relationship between the vertices, edges, and faces of a polyhedron according to Euler’s formula is F + V = E + 2.

Where F = number of faces.

V = number of vertices.

E = number of edges.

Question 3.

Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron Then verify that the relationship you found in Question 2 is valid for each polyhedron.

Answer:

We have to draw three polyhedra that are different from the platonic solids.

After drawing, all polyhedrons count the number of vertices, edges and faces of each polyhedron.

now we count vertices, edges, and faces in each solids

A triangular prism has vertices 6, edges 9 and faces 5

6 – 9 + 5 = 2

A Pentagonal prism has vertices 10, edges 15 and faces 7

10 – 15 + 7 = 2

A triangular pyramid has vertices 4, edges6 and faces

4 – 6 + 2 = 2

### Lesson 11.4 Three-Dimensional Figures

**Monitoring progress**

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 1.

Answer:

The solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.

Question 2.

Answer:

The solid have curved faces. So it is not a polyhedron.

Question 3.

Answer:

The solid is formed by polygons, so it is a polyhedron. iT has two triangles, one rectangle and two squares.

Describe the shape formed by the intersection of the plane and the solid.

Question 4.

Answer:

The cross-section is a pentagon.

Question 5.

Answer:

The cross-section is a hexagon.

Question 6.

Answer:

The cross-section is a circle.

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

### Exercise 11.4 Three-Dimensional Figures

Vocabulary and Core Concept Check

Question 1.

**VOCABULARY**

A(n) ___________ is a solid that is bounded by polygons.

Answer:

Question 2.

**WHICH ONE DOESN’T BELONG?**

Which solid does not belong with the other three? Explain your reasoning.

Answer:

Cone does not belong with the other three as it has a curved surface and others not.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, match the polyhedron with its name.

3. | A. triangular Prism |

4. | B. rectangular pyramid |

5. | C. hexagonal pyramid |

6. | D. Pentagonal prism |

Answer:

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 7.

Answer:

Question 8.

Answer:

Yes, it is a polyhedron. It is a hexagonal prism.

Question 9.

Answer:

Question 10.

Answer:

Yes, it is a polyhedron, truncated square pyramid.

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.

Question 11.

Answer:

Question 12.

Answer:

The cross-section is a square.

Question 13.

Answer:

Question 14.

Answer:

The cross-section is a hexagon

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Question 18.

Answer:

Question 19.

**ERROR ANALYSIS**

Describe and correct the error in identifying the solid.

Answer:

Question 20.

**HOW DO YOU SEE IT?**

Is the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not.

Answer:

It is an octagonal polyhedron.

In Exercises 21 – 26, sketch the polyhedron.

Question 21.

triangular prism

Answer:

Question 22.

rectangular prism

Answer:

Question 23.

pentagonal prism

Answer:

Question 24.

hexagonal prism

Answer:

Question 25.

square pyramid

Answer:

Question 26.

pentagonal pyramid

Answer:

Question 27.

**MAKING AN ARGUMENT**

Your Friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning.

Answer:

Question 28.

**ATTENDING TO PRECISION**

The figure shows a plane intersection a cube through four of its vertices. The edge length of the cube is 6 inches.

a. Describe the shape of the cross section.

Answer:

The cross-section is a rectangle.

b. What is the perimeter of the cross section?

Answer:

The perimeter is 2(l + b)

c. What is the area of the cross section?

Answer:

Area is lb.

**REASONING**

In Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

Question 29.

circle

Answer:

Question 30.

pentagon

Answer:

yes, cross-section of the cube can be a pentagon.

Question 31.

rhombus

Answer:

Question 32.

isosceles triangle

Answer:

Yes, the cross-section can be an isosceles triangle.

Question 33.

hexagon

Answer:

Question 34.

scalene triangle

Answer:

Yes, the cross-section can be scalene triangle.

Question 35.

**REASONING**

Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.

a.

b.

Answer:

Question 36.

**THOUGHT PROVOKING**

Describe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.)

Answer:

Maintaining Mathematical proficiency

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.

Question 37.

∆ABD, ∆ CDB

Answer:

Question 38.

∆JLK, ∆JLM

Answer:

∆JLK ≅ ∆JLM by SAS congruence theorem.

Question 39.

∆RQP, ∆RTS

Answer:

### 11.1 – 11.4 Quiz

Find the indicated measure.

Question 1.

m[latex]\widehat{E F}\)

Answer:

13.7 = \(\frac { m[latex]\widehat{E F}\) }{ 360 } [/latex] • 2π(7)

m\(\widehat{E F}\) = 112.13

Question 2.

arc length of \(\widehat{Q S}\)

Answer:

arc length of \(\widehat{Q S}\) = \(\frac { 83 }{ 360 } \) • 2π(4)

= 5.79

Question 3.

circumference of ⊙N

Answer:

8 = \(\frac { 48 }{ 360 } \) • 2πr

C = 60

Question 4.

Convert 26° to radians and \(\frac{5 \pi}{9}\) radians to degrees.

Answer:

26° = 26 . \(\frac { π }{ 180 } \) = \(\frac { 13π }{ 90 } \) radians

\(\frac{5 \pi}{9}\) = \(\frac{5 \pi}{9}\) . \(\frac { 180 }{ π } \) = 100°

Use the figure to find the indicated measure.

Question 5.

area of red sector

Answer:

area of red sector = \(\frac { 100 }{ 360 } \) . π(12)²

= 125.66

Question 6.

area of blue sector

Answer:

area of blue sector = \(\frac { 260 }{ 360 } \) . π(12)²

= 326.72

In the diagram, RSTUVWXY is a regular octagon inscribed in ⊙C.

Question 7.

Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:

C is center, CY is radius, CZ is apothem, ∠YCR is central angle of the polygon

Question 8.

Find m∠RCY, m∠RCZ, and m∠ZRC.

Answer:

m∠RCY = 360/8 = 45

m∠RCZ = 45/2 = 22.5

m∠ZRC = 180 – (22.5 + 90) = 67.5

Question 9.

The radius of the circle is 8 units. Find the area of the octagon.

Answer:

Area of octagon = 0.5 x 8 x 8 sin 45 = 22.62

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 10.

Answer:

It is not a polyhedron.

Question 11.

Answer:

The solid is a polyhedron. It is an octagonal pyramid.

Question 12.

Answer:

The solid is a polyhedron. It is a pentagonal prism.

Question 13.

Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.

Answer:

Question 14.

The two white congruent circles just fit into the blue circle. What is the area of the blue region?

Answer:

White circle diameter = radius of the blue circle.

6 = radius of the blue circle.

Area of blue circle = π6² = 113.09

Area of white circle = π3² = 28.27

Area of blue region = 113.09 – 2(28.27) = 56.541

Question 15.

Find the area of each rhombus tile. Then find the area of the pattern.

Answer:

Area of yellow tile = \(\frac { 1 }{ 4 } \)(15.7 x 11.4) = 44.745

area of red tile = \(\frac { 1 }{ 4 } \)(18.5 x 6) = 27.75

Area of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm

### 11.5 Volumes of Prisms and Cylinders

**Exploration 1**

Finding volume

Work with a partner: Consider a stack of square papers that is in the form of a right prism.

a. What is the volume the prism?

Answer:

The volume of the prism is B = 1/2 x h(b1 + b2)

h = 8

b1 = 2

b2 = 2

= 1/2 x 8(2 + 2)

= 1/2 x 8(4)

= 1/2 x 32

= 16 square inches.

Therefore the volume of the prism is 16 square inches.

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning.

Answer:

The volume of the prism and the twist of the slack of paper volume are the same. Because the different shapes of the prism have the same volume.

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b).

**ATTENDING TO PRECISION**

To be proficient in math, you need to communicate precisely to others.

Answer:

The conjecture is that the different shapes of the prism have the same volume but are different in surface area.

d. Use your conjecture to find the volume of the twisted stack of papers.

Answer:

The volume of the twist of the slack of paper is B = 1/2 x h(b1 + b2)

h = 8

b1 = 2

b2 = 2

= 1/2 x 8(2 + 2)

= 1/2 x 8(4)

= 1/2 x 32

= 16 square inches.

Therefore the volume of the twist and the slack of the paper is 16 square inches.

It is the same as the volume of the prism.

**Exploration 2**

Finding volume

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.

a.

Answer:

The formula for the volume of the cylinder is V = πr² h.

h = 3

r = 2

= π(2)² x 3

= π(4) x 3

= 12π

= 37.68 cu. cm

Therefore the volume of the cylinder is 37.68 cu. cm.

b.

Answer:

The formula for the volume of the cylinder is V = πr² h.

h = 15

r = 5

= π(5)² x 15

= π(25) x 15

= 375π

= 1,177.5 cu. cm

Therefore the volume of the cylinder is 1,177.5 cu. cm.

Communicate Your Answer

Question 3.

How can you find the volume of a prism or cylinder that is not a right prism or right cylinder?

Answer:

Using π we can find the volume of the prism or cylinder that is not a right prism of the right cylinder.

The cylinder and the prism have the same cross-sectional area of πr². At every level and same height.

Both the cylinder and prism have the same volume it is V = πr²h.

Question 4.

In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.

Answer:

### Lesson 11.5 Volumes of Prisms and Cylinders

**Monitoring Progress**

Find the volume of the solid.

Question 1.

Answer:

Volume = Area * height

Area = \(\frac { 1 }{ 2 } \)(5 x 9) = 22.5

Volume = 22. 5 x 8 = 180 cubic m.

Question 2.

Answer:

Area of circle = πr² = π(8)² = 64π

Volume = 64π x 14 = 2814.86 cubic ft

Question 3.

The diagram shows the dimensions of a concrete cylinder. Concrete has a density of 2.3 grams per cubic centimeter. Find the mass of the concrete cylinder to the nearest gram.

Answer:

Mass of the concrete cylinder = 32 x π(24)² = 18432π cubic in

Question 4.

**WHAT IF?**

In Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be?

Answer:

volume = lbh

60 = 5 x 3 x h

h = 4 m

Question 5.

**WHAT IF?**

In Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width.

Answer:

volume V = base x height

75 = base x 5

Base = 15 sq m

Question 6.

Prism C and prism D are similar. Find the volume of prism D.

Answer:

\(\frac { 12 }{ 3 } \) = \(\frac { 1536 }{ v } \)

v = 384

Volume of prism D = 384 cubic m

Question 7.

Find the volume of the composite solid.

Answer:

Volume = area x height

Volume = \(\frac { 1 }{ 2 } \)(10 x 3 x 6)

= 90 cubic ft

### Exercise 11.5 Volumes of Prisms and Cylinders

Vocabulary and Core Concept Check

Question 1.

**VOCABULARY**

In what type of units is the volume of a solid measured?

Answer:

Question 2.

**COMPLETE THE SENTENCE**

Density is the amount of _______ that an object has in a given unit of __________ .

Answer:

Density is the mass of the object divided by its volume.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the volume of the prism.

Question 3.

Answer:

Question 4.

Answer:

Volume V = lbh

V = 1.5 x 2 x 4 = 12 m³

Question 5.

Answer:

Question 6.

Answer:

Volume = 6 x 11 x 14

V = 924 m³

In Exercises 7 – 10. find the volume of the cylinder.

Question 7.

Answer:

Question 8.

Answer:

Volume = πr²h

V = π(13.4)² x 9.8

V = 1759.6π cm³

Question 9.

Answer:

Question 10.

Answer:

Volume V = πr²h

Shoter side = 18/2 = 9

Height h = √18² – 9² = 15.588

V = π6² x 15.588

= 1763 m³

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.

Question 11.

A prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters.

Answer:

Question 12.

A pentagonal prism has a height of 9 feet and each base edge is 3 feet.

Answer:

volume is 139.32 ft³

explanation:

Pentagon area = 15.48

Height h = 9 ft

Volume V = area x height

= 15.48 x 9 = 139.32

Question 13.

**PROBLEM SOLVING**

A piece of copper with a volume of 8.25 cubic centimeters has a mass of 73.92 grams. A piece of iron with a volume of 5 cubic centimeters has a mass of 39.35 grams. Which metal has the greater density?

Answer:

Question 14.

**PROBLEM SOLVING**

The United States has minted one-dollar silver coins called the American Eagle Silver Bullion Coin since 1986. Each coin has a diameter of 40.6 millimeters and is 2.98 millimeters thick. The density of silver is 10.5 grams per cubic centimeter. What is the mass of an American Eagle Silver Bullion Coin to the nearest grain?

Answer:

V = πr²h

V =π(20.3)² x (2.98)

V = 3856 mm³

Question 15.

**ERROR ANALYSIS**

Describe and correct the error in finding the volume of the cylinder.

Answer:

Question 16.

**ERROR ANALYSIS**

Describe and correct the error in finding the density of an object that has a mass of 24 grams and a volume of 28.3 cubic centimeters.

Answer:

Density = mass / volume

Density = \(\frac { 24 }{ 28.3 } \)

Density = 0.8480

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.

Question 17.

Volume = 500 ft^{3}

Answer:

Question 18.

Volume = 2700 yd^{3}

Answer:

Volume = 2700 yd³

12 x 5 x v = 2700

v = 15 yd

Question 19.

Volume = 80 cm^{3}

Answer:

Question 20.

Volume = 72.66 in.^{3}

Answer:

Volume = 72.66 in.^{3}

Area . x = 72.66

10.39 x = 72.66

x = 6.9 in

Question 21.

Volume = 3000 ft^{3}

Answer:

Question 22.

Volume = 1696.5 m^{3}

Answer:

Volume = 1696.5

πr²h = 1696.5

πz² x 15 = 1696.5

z² = 36.00

z = 6

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.

Question 23.

V= 154 in.^{3}, h = 11 in.

Answer:

Question 24.

V = 27 m^{3},h = 3m

Answer:

V = Bh

27 = B x 3

B = 9

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.

Question 25.

Answer:

Question 26.

Answer:

\(\frac { 12 }{ 15 } \) = \(\frac { 4608π }{ V } \)

V = 5760π

Volume of cylinder B = 5760π

In Exercises 27 and 28, the solids are similar. Find the indicated measure.

Question 27.

height x of the base of prism A

Answer:

Question 28.

height h of cylinder B

Answer:

height h of cylinder B is 40 ft

Explanation:

\(\frac { 7π }{ 5 } \) = \(\frac { 56π }{ h } \)

h = 40

In Exercises 29 – 32. find the volume of the composite solid.

Question 29.

Answer:

Question 30.

Answer:

Volume V = 89.32

Explanation:

Volume of square = 4³ = 64

Volume of semicircle = π(2)² x 4 = 8π

Question 31.

Answer:

Question 32.

Answer:

The volume of composite solid is 35 cubic ft

Explanation:

Volume of larger prism = 4 x 2 x 5 = 40

Volume of the smaller prism = 1 x 1 x 5 = 5

Volume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft

Question 33.

**MODELING WITH MATHEMATICS**

The Great Blue Hole is a cylindrical trench located off the coast of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does the Great Blue Hole contain? (1 ft^{3} ≈ 7.48 gallons)

Answer:

Question 34.

**COMPARING METHODS**

The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning.

Answer:

**REASONING**

In Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?

Question 35.

Answer:

Question 36.

Answer:

7 triangular prism candles with the given measures can be made.

Explanation:

Volume of block = 1800

The volume of triangular prism = 4 x 6 x 10 = 240

1800/240 = 7.5

Question 37.

**PROBLEM SOLVING**

An aquarium shaped like a rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fill the aquarium \(\frac{3}{4}\) fill with water. When you submerge a rock in the aquarium, the water level rises 0.25 inch.

a. Find the volume of the rock.

b. How many rocks of this size can you place in the aquarium before water spills out?

Answer:

Question 38.

**PROBLEM SOLVING**

You drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

Answer:

The density of metal is 1.4648

Explanation:

Density = \(\frac { Mass }{ Volume } \)

Volume V = 4.8 x 64 = 307.2

Density = \(\frac { 450 }{ 307.2 } \) = 1.4648

Question 39.

**WRITING**

Both of the figures shown arc made up of the same number of congruent rectangles. Explain how Cavalieri’s Principle can be adapted to compare the areas of these figures.

Answer:

Question 40.

**HOW DO YOU SEE IT?**

Each stack of memo papers contains 500 equally-sized sheets of paper. Compare their volumes. Explain your reasoning.

Answer:

Given that,

Each stack of memo papers contains 500 equally-sized sheets of paper.

The sheets of paper are in the shape of a square prism.

The different shapes of the prism have the same volume.

Question 41.

**USING STRUCTURE**

Sketch the solid formed by the net. Then find the volume of the solid.

Answer:

Question 42.

**USING STRUCTURE**

Sketch the solid with the given views. Then find the volume of the solid.

Answer:

Volume = 2.5 x 3.5 x 6

Volume = 52.5

Question 43.

**OPEN-ENDED**

Sketch two rectangular prisms that have volumes of 1000 cubic inches hut different surface areas. Include dimensions in your sketches.

Answer:

Question 44.

**MODELING WITH MATHEMATICS**

Which box gives you more cereal for your money? Explain.

Answer:

First one gives more cerel for your money.

Explanation:

Bigger one volume = 16 x 4 x 10 = 640

Smaller one volume = 2 x 8 x 10 = 160

6 – 640 means 1 – 106.66

2 – 160 means 1 – 80

Question 45.

**CRITICAL THINKING**

A 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.

Answer:

Question 46.

**CRITICAL THINKING**

The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.

Answer:

Let the height of cylinder X be h, radius be r and its volume is πr²h

So, the height of cylinder Y is h/2 and radius is 2r, then the volume is 2πr²h

From both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.

Question 47.

**USING STRUCTURE**

Find the volume of the solid shown. The bases of the solid are sectors of circles.

Answer:

Question 48.

**MATHEMATICAL CONNECTIONS**

You drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.

Answer:

r = √R²/2

Explanation:

The radius of a solid cylinder without a hole is R. So its volume is πR²h

As per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is πR²h/2

Volume of cylinder V = πr²h

πR²h/2 = πr²h

R²/2 = r²

r = √R²/2

r = \(\frac { R√2 }{ 2 } \)

Question 49.

**ANALYZING RELATIONSHIPS**

How can you change the height of a cylinder so that the volume is increased by 25% but the radius remains the same?

Answer:

Question 50.

**ANALYZING RELATIONSHIPS**

How can you change the edge length of a cube so that the volume is reduced by 40%?

Answer:

Write the equation of volume of rectangular prism which can be used to evaluate the cube volume

Volume = s x s x s

The above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.

Question 51.

**MAKING AN ARGUMENT**

You have two objects of equal V0lume. Your friend says you can compare the densities of the objects by comparing their mass, because the heavier object will have a greater density. Is your friend correct? Explain your reasoning.

Answer:

Question 52.

**THOUGHT PROVOKING**

Cavalieri’s Principle states that the two solids shown below have the same volume. Do they also have the same surface area? Explain your reasoning.

Answer: Cavalieri’s Principle states that the two solids shown have the same volume. They have different surface areas because the first solid have vertical side surfaces whereas the second solid has slant side surfaces. The slant surface has a greater surface area than the vertical surface.

Question 53.

**PROBLEM SOLVING**

A barn is in the shape of a pentagonal prism with the dimensions shown, The volume of the barn is 9072 cubic feel. Find the dimensions of each half of the root.

Answer:

Question 54.

**PROBLEM SOLVING**

A wooden box is in the shape of a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round your answer to the nearest tenth.

Answer:

Pentagon area = \(\frac { 5 }{ 2 } \)(4)² sin 72

A = 38

Volume = Area x height

Volume = 38 x 6 = 228

Maintaining Mathematical Proficiency

Find the surface area of the regular pyramid.

Question 55.

Answer:

Question 56.

Answer:

Surface area = base area + 3bs

Surface area = 166.3 + 3(8)(10)

= 406.3

Question 57.

Answer:

### 11.6 Volumes of Pyramids

**Exploration 1**

Finding the Volume of a Pyramid

Work with a partner: The pyramid and the prism have the same height and the same square base.

When the pyramid is filled with sand and poured into the prism, it takes three pyramids to fill the prism.

Use this information to write a formula for the volume V of a pyramid.

**LOOKING FOR STRUCTURE**

To be proficient in math, you need to look closely to discern a pattern or structure.

Answer:

Given,

The area of the square base of the pyramid is equal to that of the prism and height are same of both the bodies and also volume of prism = 3 × volume of pyramid

Ab × h = 3V

V = Ab . h/3

where Ab = Area of the square pyramid as well as prism

h = height of the prism

**Exploration 2**

Finding the Volume of a Pyramid

Work with a partner: Use the formula you wrote in Exploration 1 to find the volume of the hexagonal pyramid.

Answer:

V = √3/2 × a² × h

V = √3/2 × 4 × 3

V = 10.39 cubic. in

Communicate Your Answer

Question 3.

How can you find the volume of a pyramid?

Answer: The volume of a pyramid is found using the formula V = (1/3) Bh, where ‘B’ is the base area and ‘h’ is the height of the pyramid. As we know the base of a pyramid is any polygon, we can apply the area of polygons formulas to find ‘B’.

Question 4.

In Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning.

Answer:

### Lesson 11.6 Volumes of Pyramids

**Monitoring Progress**

Find the volume of the pyramid.

Question 1.

Answer:

The volume of the pyramid is 400 cm³

Explanation:

Volume V = \(\frac { 1 }{ 3 } \)Bh

V = \(\frac { 1 }{ 3 } \)(10 x 10 x 12)

V = 400

Question 2.

Answer:

The volume of the pyramid is 2494.13 cm³

Explanation:

Volume V = \(\frac { 1 }{ 3 } \)Bh

V = \(\frac { 1 }{ 3 } \)(374.12 x 20)

V = 2494.13

Question 3.

The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.

Answer:

The side length of the square base is 5 m

Explanation:

Volume V = \(\frac { 1 }{ 3 } \)Bh = 75

\(\frac { 1 }{ 3 } \)B(9) = 75

B = 25

s = 5

Question 4.

Find the height of the triangular pyramid at the left.

Answer:

The height of the triangular pyramid is 8 m

Explanation:

V = 24

\(\frac { 1 }{ 3 } \)Bh = 24

B = \(\frac { 1 }{ 2 } \)(3 x 6) = 9

\(\frac { 1 }{ 3 } \)(9)h = 24

h = 8

Question 5.

Pyramid C and pyramid D are similar. Find the volume of pyramid D.

Answer:

The volume of pyramid D is 12 m³

Explanation:

\(\frac { volume of pyramid C }{ volume of pyramid D } \) = (\(\frac { pyramid C base }{ pyramid D base } \))³

\(\frac { 324 }{ V } \) = (\(\frac { 9 }{ 3 } \))³

V = 12

Question 6.

Find the volume of the composite solid.

Answer:

the volume of solid = 96

Explanation:

Volume of prism = Bh

B = 8 x 2 = 16

V = 16 x 5 = 80

\(\frac { 1 }{ 3 } \)Bh

= \(\frac { 1 }{ 3 } \)(16 x 3) = 16

the volume of solid = 16 + 80 = 96

### Exercise 11.6 Volumes of Pyramids

Vocabulary and Core Concept Check

Question 1.

**VOCABULARY**

Explain the difference between a triangular prism and a triangular pyramid.

Answer:

Question 2.

**REASONING**

A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.

Answer:

Square pyramid = 1/3 Bh

Cube = BH

So, the volume of the square pyramid is 1/3 of the volume of the cube.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the volume of the pyramid.

Question 3.

Answer:

Question 4.

Answer:

V = 6 in³

Explanation:

V = \(\frac { 1 }{ 3 } \)Bh

B = 2 x 3 = 6

V = \(\frac { 1 }{ 3 } \)(6 x 3)

In Exercises 5 – 8, find the indicated measure.

Question 5.

A pyramid with a square base has a volume of 120 cubic meters and a height of 10 meters. Find the side length of the square base.

Answer:

Question 6.

A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.

Answer:

The side length of the square base is 12 ft

Explanation:

A pyramid with a square base has a volume of 912 cubic feet

h = 19

\(\frac { 1 }{ 3 } \)Bh = 912

\(\frac { 1 }{ 3 } \)B(19) = 912

B = 144

s = 12

Question 7.

A pyramid with a rectangular base has a volume of 480 cubic inches and a height of 10 inches. The width of the rectangular base is 9 inches. Find the length of the rectangular base.

Answer:

Question 8.

A pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.

Answer:

The width of the rectangular base is 3 cm

Explanation:

A pyramid with a rectangular base has a volume of 105 cubic centimeters

h = 15

l = 7

\(\frac { 1 }{ 3 } \)Bh = 105

\(\frac { 1 }{ 3 } \)lbh = 105

\(\frac { 1 }{ 3 } \)(7 x 15 x b) = 105

b = 3

Question 9.

**ERROR ANALYSIS**

Describe and correct the error in finding the volume of the pyramid.

Answer:

Question 10.

**OPEN-ENDED**

Give an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning.

Answer:

Let the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units

Therefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1/3 x 4 x 2 x 15 = 40 cubic units

In Exercises 11 – 14, find the height of the pyramid.

Question 11.

Volume = 15 ft^{3}

Answer:

Question 12.

Volume = 224 in.^{3}

Answer:

The height of the pyramid is 10.5 in

Explanation:

Volume = 224

\(\frac { 1 }{ 3 } \)Bh = 224

B = 8² = 64

\(\frac { 1 }{ 3 } \)(64)h = 224

h = 10.5

Question 13.

Volume = 198 yd^{3}

Answer:

Question 14.

Volume = 392 cm^{3}

Answer:

The height of the pyramid is 12 cm

Explanation:

Volume = 392

\(\frac { 1 }{ 3 } \)Bh = 392

B = 14 x 7 = 98

\(\frac { 1 }{ 3 } \)(98)h = 392

h = 12

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.

Question 15.

Answer:

Question 16.

Answer:

Volume of A = 80

Explanation:

\(\frac { Volume of B }{ Volume of A } \) = (\(\frac { Side of B }{ side of A } \))³

\(\frac { V }{ 10 } \) = (\(\frac { 6 }{ 3 } \))³

V = 8 x 10

Volume of A = 80

In Exercises 17 – 20, find the volume of the composite solid.

Question 17.

Answer:

Question 18.

Answer:

Composite solid volume = 306

Explanation:

Base area = \(\frac { 1 }{ 2 } \)bh = \(\frac { 1 }{ 2 } \)(12 x 9) = 54

Bottom solid volume V = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(54 x 10)

V = 180

Top solid volume v = \(\frac { 1 }{ 3 } \)(54 x 7) = 126

Composite solid volume = 180 + 126 = 306

Question 19.

Answer:

Question 20.

Answer:

Composite solid volume = 1152

Explanation:

Volume of Box = 12 x 12 x 12 = 1728

Square pyramid volume = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(144 x 12)

= 576

Composite solid volume = 1728 – 576 = 1152

Question 21.

**ABSTRACT REASONING**

A pyramid has a height of 8 feet and a square base with a side length of 6 feet.

a. How does the volume of the pyramid change when the base slays the same and the height is doubled?

b. How does the volume of the pyramid change when the height stays the same and the side length of the base is doubled?

C. Are your answers 10 parts (a) and (b) true for any square pyramid? Explain your reasoning.

Answer:

Question 22.

**HOW DO YOU SEE IT?**

The cube shown is formed by three pyramids. each with the same square base and the same height. How could you use this to verify the formula for the volume of a pyramid?

Answer:

Question 23.

**CRITICAL THINKING**

Find the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram. m∠ABC = 35°

Answer:

Question 24.

**THOUGHT PROVOKING**

A frustum of a pyramid is the part of the pyramid that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a square pyramid in terms of a, b, and h. (Hint: Consider the “missing” top of the pyramid and use similar triangles.)

Answer:

The frustum of a square pyramid is shown in the figure. The question requires creating a formula for the volume of the frustum. To create the required formula, draw the complete pyramid by creating the missing top on the frustum. Then, using similar triangles, find the total height of the pyramid in terms of h, b, and a. Finally, write an equation for the volume of the frustum and use the formula for the volume of the pyramid to create the formula.

Question 25.

**MODELING WITH MATHEMATICS**

Nautical deck prisms were used as a safe way to illuminate decks on ships. The deck prism shown here is composed of the following three solids: a regular hexagonal prism with an edge length of 3.5 inches and a height of 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches arid a height of 0.25 inch, and a regular hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. Find the volume of the deck prism.

Answer:

Maintaining Mathematical Proficiency

Find the value of X. Round your answer to the nearest tenth.

Question 26.

Answer:

tan 35 = \(\frac { 9 }{ x } \)

0.7 = \(\frac { 9 }{ x } \)

x = 12.8

Question 27.

Answer:

Question 28.

Answer:

tan 30 = \(\frac { x }{ 10 } \)

0.577 = \(\frac { x }{ 10 } \)

x = 5.77

Question 29.

Answer:

### 11.7 Surface Areas and Volumes of Cones

**Exploration 1**

Finding the Surface Area of a Cone

Work with a partner: Construct a circle with a radius of 3 inches. Mark the circumference of the circle into six equal parts, and label the length of each part. Then cut out one sector of the circle and make a cone.

a. Explain why the base of the cone is a circle. What are the circumference and radius

of the base?

Answer:

b. What is the area of the original circle? What is the area with one sector missing?

Answer:

c. Describe the surface area of the cone, including the base. Use your description to find the surface area.

Answer:

**Exploration 2**

Finding the Volume of a Cone

Work with a partner: The cone and the cylinder have the same height and the same circular base.

When the cone is filled with sand and poured into the cylinder. it takes three cones to fill the cylinder.

Use this information to write a formula for the volume V of a cone.

**CONSTRUCTING VIABLE ARGUMENTS**

To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.

Answer:

Communicate Your Answer

Question 3.

How can you find the surface area and the volume of a cone?

Answer:

Question 4.

In Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern.

Answer:

### Lesson 11.7 Surface Areas and Volumes of Cones

Monitoring progress

Question 1.

Find the surface area of the right cone.

Answer:

The surface area of the right cone is 436.17 m²

Explanation:

r = 7.8

l = 10

S = πr² + πrl

S = π(7.8)² + π(7.8 x 10)

S = 436.17

Find the volume of the cone.

Question 2.

Answer:

The volume of the cone is 2206.44 in³

Explanation:

r = 7, h = 13

l = √13² – 7²= 10.95

S = πr² + πrl

S = π7² + π(7 x 10.95)

S = 394.74

Volume V = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 7² x 13)

V = 2206.44

Question 3.

Answer:

The volume of the cone is 163.4 m³

Explanation:

h = √8² – 5² = 6.24

Volume V = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 5² x 6.24)

V = 163.4

Question 4.

Cone C and cone D are similar. Find the volume of cone D.

Answer:

Volume of cone D = 18.84 cm³

Explanation:

\(\frac { Volumeof cone C }{ Volume of cone D } \) = (\(\frac { height of C }{ height of D } \))³

\(\frac { 384π }{ Volume of cone D } \)= (\(\frac { 8 }{ 2 } \))³

Volume of cone D = 18.84

Question 5.

Find the volume of the composite solid.

Answer:

Composite solid volume = 329.86 cm³

Explanation:

Volume of cylinder = πr²h = π(3)² x 10 = 90π

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

= \(\frac { 1 }{ 3 } \)(π x 3² x 5)

= 15π

Composite solid volume = 15π + 90π = 105π

### Exercise 11.7 Surface Areas and Volumes of Cones

Vocabulary and Core Concept Check

Question 1.

**WRITING**

Describe the differences between pyramids and cones. Describe their similarities.

Answer:

Question 2.

**COMPLETE THE SENTENCE**

The volume of a cone with radius r and height h is \(\frac{1}{3}\) the volume of a(n) __________ with radius r and height h.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the right cone.

Question 3.

Answer:

Question 4.

Answer:

The surface area of cone is 219.44 sq cm.

Explanation:

S = πr² + πrl

S = π(5.5)² + π(5.5 x 7.2)

S = 219.44

Question 5.

A right cone has a radius of 9 inches and a height of 12 inches.

Answer:

Question 6.

A right cone has a diameter of 11.2 feet and a height of 19.2 feet.

Answer:

The surface area is 421.52 sq ft.

Explanation:

r = 5.6

h = 19.2

l = √19.2² – 5.6² = 18.36

Surface area S = πr² + πrl

S = π(5.6)² + π(5.6 x 18.36)

S = 421.52

In Exercises 7 – 10, find the volume of the cone.

Question 7.

Answer:

Question 8.

Answer:

The volume is 2.09 cubic meter

Explanation:

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π(1)² x 2)

V = 2.09

Question 9.

A cone has a diameter of 11.5 inches and a height of 15.2 inches.

Answer:

Question 10.

A right cone has a radius of 3 feet and a slant height of 6 feet.

Answer:

The volume is 56.54 cubic ft

Explanation:

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 3² x 6)

V = 56.54

In Exercises 11 and 12, find the missing dimension(s).

Question 11.

Surface area = 75.4 cm^{2}

Answer:

Question 12.

Volume = 216π in.^{3}

Answer:

The radius is 6.13 in

Explanation:

Volume = 216π in.^{3}

\(\frac { 1 }{ 3 } \)(πr²h) = 216

\(\frac { 1 }{ 3 } \)(πr² x 18) = 216

r = 6.13

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.

Question 13.

Answer:

Question 14.

Answer:

Volume of cone B = 24.127

Explanation:

\(\frac { Volumeof cone A }{ Volume of cone B } \) = (\(\frac { height of A }{ height of B } \))³

\(\frac { 120π }{ Volume of cone B } \) = (\(\frac { 10 }{ 4 } \))³

Volume of cone B = 24.127

In Exercises 15 and 16, find the volume of the composite solid.

Question 15.

Answer:

Question 16.

Answer:

Volume of the composite solid = 97.93 cubic m

Explanation:

Volume of box = lbh

V = 51 x 5.1 x 5.1 = 132.651

Cone volume = \(\frac { 1 }{ 3 } \)(πr²h)

v = \(\frac { 1 }{ 3 } \)(π x 2.55² x 5.1)

v = 34.72

Volume of the composite solid = 132.651 – 34.72 = 97.93

Question 17.

**ANALYZING RELATIONSHIPS**

A cone has height h and a base with radius r. You warn to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain.

Answer:

Question 18.

**HOW DO YOU SEE IT**

A snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform any calculations.

a. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a large container? Explain.

Answer:

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 3² x 8) = 75.39

Volume of cylinder = πr²h = π x 3² x 8 = 226.19

Volume of cylinder / Volume of cone = \(\frac { 226.19 }{ 75.39 } \) = 3

You have to buy 3 small containers of popcorn to equal the amount of popcorn in a large container.

b. Which container gives you more popcorn for your money? Explain.

Answer:

$1.25 -> 75.39 i.e $1 = 60.312

$2.50 -> 226.19 i.e $1 = 90.47

So, large containers gives you more popcorn for your money

In Exercises 19 and 20. find the volume of the right cone.

Question 19.

Answer:

Question 20.

Answer:

Volume of cone is 575.62 cubic yd

Explanation:

tan 32 = \(\frac { 7 }{ h } \)

h = 11.21

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 7² x 11.21)

V = 575.62

Question 21.

**MODELING WITH MATHEMATICS**

A cat eats hail a cup of food, twice per day. Will the automatic pet feeder hold enough food for 10 days? Explain your reasoning. (1 cup ≈ 14.4 in.^{3})

Answer:

Question 22.

**MODELING WITH MATHEMATICS**

During a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm^{3})

Answer:

17.45 seconds

Explanation:

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 5² x 10)

V = 261.8

\(\frac { 261.8 }{ 15 } \) = 17.45

Question 23.

**REASONING**

To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m∠ABC.

Answer:

Question 24.

**THOUGHT PROVOKING**

A frustum of a cone is the part of the cone that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the “missing” top of the cone and use similar triangles.)

Answer:

Volume V = (1/3) * π * h * (r1² + r2² + (r1 * r2))

Question 25.

**MAKING AN ARGUMENT**

In the figure, the two cylinders are congruent The combined height of the two smaller cones equals the height of the larger cone. Your friend claims that this means the total volume of the two smaller cones is equal to the volume of the larger cone. Is your friend correct? Justify your answer.

Answer:

Question 26.

**CRITICAL THINKING**

When the given triangle is rotated around each of its sides. solids of revolution are formed. Describe the three solids and find their volumes. Give your answers in terms of π.

Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 27.

area of a circle with a radius of 7 feet

Answer:

Question 28.

area of a circle with a diameter of 22 centimeters

Answer:

d = 11

A = πr²

A = 121π

Question 29.

diameter of a circle with an area of 256 square meters

Answer:

Question 30.

radius of a circle with an area of 529 π square inches

Answer:

A = πr²

529π = πr²

r = 23

### 11.8 Surface Areas and Volumes of Spheres

**Exploration 1**

Finding the Surface Area of a Sphere

Work with a partner: Remove the covering from a baseball or softball.

You will end up with two “figure 8” pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball.

Use the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.

**USING TOOLS STRATEGICALLY**

To be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.

Answer:

**Exploration 2**

Finding the volume of a sphere

Work with a partner: A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.

When half of the sphere (a hemisphere) is filled with sand and poured into the cylinder, it takes three hemispheres to till the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.

Answer:

Communicate Your Answer

Question 3.

How can you find the surface area and the volume of a sphere?

Answer:

Question 4.

Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters.

Answer:

### Lesson 11.8 Surface Areas and Volumes of Spheres

**Monitoring Progress**

Find the surface area of the sphere.

Question 1.

Answer:

The surface area of the sphere is 5026.54 ft²

Explanation:

D = 40

r = 20

The surface area of the sphere = 4πr²

S = 4 x π x (20)²

S = 5026.54 ft²

Question 2.

Answer:

The surface area of the sphere is 113.09 ft²

Explanation:

Circumference C = 6π

2πr = 6π

r = 3

The surface area of the sphere = 4πr²

S = 4π x 3²

S = 113.09

Question 3.

Find the radius of the sphere.

Answer:

The radius of the sphere is 2.73 m

Explanation:

The surface area of the sphere = 4πr²

30π = 4πr²

r = 2.73

Question 4.

The radius of a sphere is 5 yards. Find the volume of the sphere.

Answer:

The volume of the sphere is 523.59 yards³

Explanation:

r = 5

The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 5³

V = 523.59 yards³

Question 5.

The diameter of a sphere is 36 inches. Find the volume of the sphere.

Answer:

The volume of the sphere is 24429.02 in³

Explanation:

D = 36

r = 18

The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 18³

V = 24429.02

Question 6.

The surface area of a sphere is 576π square centimeters. Find the volume of the sphere.

Answer:

The volume of the sphere is 2304π cm³

Explanation:

The surface area of the sphere = 4πr²

576π = 4πr²

r = 12

The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 12³

V = 2304π

Question 7.

Find the volume of the composite solid at the left.

Answer:

The volume of the composite solid is 7.324 m³

Explanation:

The volume of cone = πr²\(\frac { h }{ 3 } \)

= π x 1² x \(\frac { 5 }{ 3 } \) = 5.23

The volume of sphere = \(\frac { 4 }{ 3 } \)πr³

= \(\frac { 4 }{ 3 } \)π x 1³ = 4.188

The volume of the composite solid = The volume of cone + The volume of sphere/2

= 5.23 + 4.188/2

= 7.324 m³

### Exercise 11.8 Surface Areas and Volumes of Spheres

Question 1.

**VOCABULARY**

When a plane intersects a sphere. what must be true for the intersection to be a great circle?

Answer:

Question 2.

**WRITING**

Explain the difference between a sphere and a hemisphere.

Answer:

Hemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \(\frac { 4 }{ 3 } \)πr³ and hemisphere volume is \(\frac { 2 }{ 3 } \)πr³. The surface area of the sphere is 4πr² and hemisphere surface area is 3πr².

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the sphere.

Question 3.

Answer:

Question 4.

Answer:

The surface area of the sphere is 225π cm²

Explanation:

The surface area of the sphere = 4πr²

S = 4π x 7.5²

S = 225π

Question 5.

Answer:

Question 6.

Answer:

The surface area of the sphere is 8π ft²

Explanation:

C = 4π

2πr = 4π

r = 2

The surface area of the sphere = 4πr²

S = 4π x 2²

S = 8π

In Exercises 7 – 10. find the indicated measure.

Question 7.

Find the radius of a sphere with a surface area of 4π square feet.

Answer:

Question 8.

Find the radius of a sphere with a surface area of 1024π square inches.

Answer:

The radius of a sphere is 16 in

Explanation:

The surface area of the sphere = 1024π

4πr² = 1024π

r = 16

Question 9.

Find the diameter of a sphere with a surface area of 900π square meters.

Answer:

Question 10.

Find the diameter of a sphere with a surface area of 196π square centimeters.

Answer:

The diameter of a sphere is 14 cm

Explanation:

The surface area of the sphere = 196π

4πr² = 196π

r = 7

D = 2(7) = 14

In Exercises 11 and 12, find the surface area of the hemisphere.

Question 11.

Answer:

Question 12.

Answer:

The surface area of the hemisphere is 108π in²

Explanation:

D = 12, r = 6

The surface area of the sphere = 3πr²

S = 3π x 6²

S = 108π

In Exercises 13 – 18. find the volume of the sphere.

Question 13.

Answer:

Question 14.

Answer:

The volume of the sphere is 268.08 ft³

Explanation:

r = 4 ft

Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 4³

V = 268.08 ft

Question 15.

Answer:

Question 16.

Answer:

The volume of the sphere is 1436.75 ft³

Explanation:

D = 14 ft

r = 7 ft

Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 7³

V = 1436.75 ft

Question 17.

Answer:

Question 18.

Answer:

The volume of the sphere is 179.89 in³

Explanation:

C = 7π

2πr = 7π

r = 3.5

Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 3.5³

V = 179.89 in

In Exercises 19 and 20, find the volume of the sphere with the given surface area.

Question 19.

Surface area = 16π ft^{2}

Answer:

Question 20.

Surface area = 484π cm^{2}

Answer:

The volume of the sphere is 5575.27 cm³

Explanation:

Surface area = 484π

4πr² = 484π

r = 11

Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 11³

V = 5575.27

Question 21.

**ERROR ANALYSIS**

Describe and correct the error in finding the volume of the sphere.

Answer:

Question 22.

**ERROR ANALYSIS**

Describe and correct the error in finding the volume of the sphere.

Answer:

Diameter = 3

radius = 1.5

Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x (1.5)³

V = 14.137 cubic in

In Exercises 23 – 26, find the volume of the composite solid.

Question 23.

Answer:

Question 24.

Answer:

Volume is 288π ft³

Explanation:

Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³

= \(\frac { 2 }{ 3 } \)π x 6³ = 144π

volume of the cone = πr²\(\frac { h }{ 3 } \)

= π x 6² x \(\frac { 12 }{ 3 } \) = 144π

Area of circle = πr² = π x 6² = 36π

Volume of hemipshere + volume of the cone = 144π + 144π = 288π

Question 25.

Answer:

Question 26.

Answer:

The volume of solid is 296π m³

Explanation:

Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³

= \(\frac { 2 }{ 3 } \)π x 6³ = 144π

Volume of cylinder = πr²h

= π x 6² x 14 = 504π

Volume of solid = 504π – 2(144π) = 296π

In Exercises 27 – 32, find the surface area and volume of the ball.

Question 27.

bowling ball

d = 8.5 in.

Answer:

Question 28.

basketball

C = 29.5 in.

Answer:

The surface area is 277 in², volume is 43212.27 in³

Explanation:

C = 29.5

2πr = 29.5

r = 4.69

Surface area = 4πr²

S = 4π x 4.69² = 277

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 4.69³

V = 43212.27

Question 29.

softball

C = 12 in.

Answer:

Question 30.

golf ball

d = 1.7 in.

Answer:

The surface area is 9.07 in², volume is 2.57 in³

Explanation:

d = 1.7

r = 0.85

Surface area = 4πr²

S = 4π x 0.85² = 9.07

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 0.85³

V = 2.57

Question 31.

volleyball

C = 26 in.

Answer:

Question 32.

baseball

C = 9 in.

Answer:

The surface area is 25.78 in², volume is 12.24 in³

Explanation:

C = 9

2πr = 9

r = 1.43

Surface area = 4πr²

S = 4π x 1.43²

S = 25.78

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 1.43³

V = 12.24

Question 33.

**MAKING AN ARGUMENT**

You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is our friend correct? Explain your reasoning.

Answer:

Question 34.

**REASONING**

A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.

Answer:

The surface area is 1018 in², volume is 3054.02 in³

Explanation:

Diameter = 18

radius r = 9

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 9³

V = 3054.02

Surface area = 4πr²

S = 4π x 9²

S = 1018

Question 35.

**MODELING WITH MATHEMATICS**

A silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.

Answer:

Question 36.

**MODELING WITH MATHEMATICS**

Three tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches.

a. Find the volume of a tennis ball.

Answer:

C = 8 in

2πr = 8

r = 1.27

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 1.27³ = 8.64

The volume of tennis ball = 8.64 in³

b. Find the amount of space within the cylinder not taken up by the tennis balls.

Answer:

The surface area of tennis ball S = 4πr²

S = 4π x 1.27² = 20.26

Area of cylinder s = 2πrh+2πr²

s = 2π x 1.43 x 8+2π x 1.43²

s = 84.72

Remaining space = 84.72 – 20.26 = 64.46 in²

Question 37.

**ANALYZING RELATIONSHIPS**

Use the table shown for a sphere.

a. Copy and complete the table. Leave your answers in terms of π.

b. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled?

c. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?

Answer:

Question 38.

**MATHEMATICAL CONNECTIONS**

A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.

Answer:

x =11

Explanation:

Surface area = 4πr²

784π = πr²

r = 28

2r = diameter = 4(x + 3)

r = 2(x + 3)

28 = 2(x + 3)

x = 11

Question 39.

**MODELING WITH MATHEMATICS**

The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles.

a. Find the surface area of Earth and the moon.

b. Compare the surface areas of Earth and the moon.

c. About 70% of the surface of Earth is water. How many square miles of water are on Earth’s surface?

Answer:

Question 40.

**MODELING WITH MATHEMATICS**

The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.

a. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.)

Answer:

Surface area of cylinder = 2πrh

S = 2π x 3960 x 3250 = 80875080

surface area of earth = 4πr²

= 4π x 3960² = 197086348.8

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.

Answer:

Probability of meteorites hitting the torrid zone = 80875080/197086348.8 = 0.4104

Question 41.

**ABSTRACT REASONING**

A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning.

Answer:

Question 42.

**HOW DO YOU SEE IT?**

The formula for the volume of a hemisphere and a Cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning.

Answer:

The hemisphere has the highest volume.

Explanation:

Volume of hemisphere v = \(\frac { 2 }{ 3 } \)πr³

Volume of cone V = \(\frac { 1 }{ 3 } \)πr²h

If r = h

Volume of cone V = \(\frac { 1 }{ 3 } \)πr² x r = \(\frac { 1 }{ 3 } \)πr³

So, the hemisphere has the highest volume.

Question 43.

**CRITICAL THINKING**

Let V be the volume of a sphere. S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. (Hint: Start with the ratio \(\frac{V}{S}\).)

Answer:

Question 44.

**THOUGHT PROVOKING**

A spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.

Answer:

The surface area of a spherical lune is 2θ R², where R is the radius of the sphere and θ is the dihedral angle in radians between the two half-great circles.

Question 45.

**CRITICAL THINKING**

The volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder.

Answer:

Question 46.

**PROBLEM SOLVING**

A spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = \(\frac{\pi h}{6}\) (3a^{2} + h^{2}), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to find the volume of each spherical cap.

a. r = 5ft, a = 4ft

Answer:

The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²).

Where a is the radius

h is the height of the cap.

Where a = r = 5ft

h = 4ft

V = π(4)/6 x (3(5)² +(4)²)

= π(4)/6 x 91

= 4π/6 x 91

= 12.56/6 x 91

Therefore the volume of the spherical cap = 190.49 cu. feet.

b. r = 34 cm, a = 30 cm

Answer:

The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²).

Where a is the radius

h is the height of the cap.

Where a = r = 34cm

h = 30cm

V = π(30)/6 x (3(34)² +(30)²)

= π(30)/6 x (3(1156) + (900))

= π(30)/6 x 6,168

= 94.2/6 x 6,168

Therefore the volume of the spherical cap is 96,837.6 cu. cm.

c. r = 13 m, h = 8 m

Answer:

The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²).

Where a is the radius

h is the height of the cap.

Where a = r = 13cm

h = 8cm

V = π(8)/6 x (3(13)² +(8)²)

= 8π/6 x (3(169 + 64))

= 8π/6 x (699)

= 25.12/6 x 699

Therefore the volume of the spherical cap is 2,926.48 cu. cm

d. r=75 in., h = 54in.

Answer:

The formula for the volume of the spherical cap is V = πh/6 x (3a² +h²).

Where a is the radius

h is the height of the cap.

Where r = 75in

h = 54in

V = π(54)/6 x (3(75)² +(54)²)

= 54π/6 x (3(5,625 + 2,916)

= 54π/6 x (25,623)

= 169.56π/6 x 25,623

= 532.4184 x 25,623

Therefore the volume of the spherical cap is 13,642,156.66 cu. in.

Question 47.

**CRITICAL THINKING**

A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.

Answer:

Maintaining Mathematical Proficiency

Solve the triangle. Round decimal answers to the nearest tenth.

Question 48.

A = 26°, C = 35°, b = 13

Answer:

B = 119°, a = 7.16, c = 9.5

Explanation:

B = 180 – (26 + 35) = 119

\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)

\(\frac { sin 26 }{ a } \) = \(\frac { sin 119 }{ 13 } \)

a = 7.16

\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)

\(\frac { sin 35 }{ c } \) = \(\frac { sin 119 }{ 13 } \)

c = 9.5

Question 49.

B = 102°, C = 43°, b = 21

Answer:

Question 50.

a = 23, b = 24, c = 20

Answer:

A = 62.2, B = 65.5, C = 49.4

Explanation:

a² = b² + c² – 2bc cos A

23² = 24²+ 20² – 2(24 x 20) cos A

A = 62.2

\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin B }{ 24 } \)

B = 65.5

\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin C }{ 20 } \)

C = 49.4

Question 51.

A = 103°, b = 15, c = 24

Answer:

### Circumference, Area, and Volume Review

### 11.1 Circumference and Arc Length

Find the indicated measure.

Question 1.

diameter of ⊙P

Answer:

diameter of ⊙P is 29.99

Explanation:

Circumference = 94.24

πd = 94.24

d = 29.99

Question 2.

circumference of ⊙F

Answer:

circumference of ⊙F = 56.57

Explanation:

5.5 = \(\frac { 35 }{ 360 } \) . C

C = 56.57

Question 3.

arc length of \(\widehat{A B}\)

Answer:

arc length of \(\widehat{A B}\) = 26.09

Explanation:

arc length of \(\widehat{A B}\) = \(\frac { 115 }{ 360 } \) . 2π(13)

= 26.09

Question 4.

A mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?

Answer:

The tire travels 2613.80 inches.

Explanation:

D = 26 in

r = 13 in

Circumference C = 2π(13) = 81.68

32 revolutions = 32 x 81.68 = 2613.80

### 11.2 Areas of Circles and Sectors

Find the area of the blue shaded region.

Question 5.

Answer:

Area = \(\frac { 240 }{ 360 } \) . π(9)²

= 169.64

Question 6.

Answer:

Area of shaded region = 11.43

Explanation:

Area of rectangle = 6 x 4 = 24

Area of semicircle = π(2)² = 4π

Area of shaded region = 24 – 4π = 11.43

Question 7.

Answer:

Area of shaded region = 173.13

Explanation:

Area of small region = 27.93 = \(\frac { 50 }{ 360 } \) . πr²

πr² = 201.096

r = 8

Area of shaded region = \(\frac { 310 }{ 360 } \) . π(8)²

= 173.13

### 11.3 Areas of Polygons

Find the area of the kite or rhombus.

Question 8.

Answer:

Area = 65

Explanation:

Area = \(\frac { 1 }{ 4 } \)(d₁d₂)

A = \(\frac { 1 }{ 4 } \)(13 x 20)

A = 65

Question 9.

Answer:

Area = 48

Explanation:

Area = \(\frac { 1 }{ 4 } \)(d₁d₂)

A = \(\frac { 1 }{ 4 } \)(16 x 12)

A = 48

Question 10.

Answer:

Area = 52.5

Explanation:

Area = \(\frac { 1 }{ 4 } \)(d₁d₂)

A = \(\frac { 1 }{ 4 } \)(14 x 15)

A = 52.5

Find the area of the regular polygon.

Question 11.

Answer:

Area = \(\frac { 3√3 }{ 2 } \)a²

A = \(\frac { 3√3 }{ 2 } \)(8.8)²

A = 201.195

Question 12.

Answer:

Area = \(\frac { 1 }{ 2 } \)(n . a. s)

A = \(\frac { 1 }{ 2 } \)(9 . 5.2 . 7.6)

A = 117.84

Question 13.

Answer:

Area = \(\frac { 1 }{ 2 } \)(n . a. s)

A = \(\frac { 1 }{ 2 } \)(5 . 4 . 3.3)

A = 33

Question 14.

A platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.

Answer:

Area = \(\frac { 1 }{ 2 } \)(n . a. s)

A = \(\frac { 1 }{ 2 } \)(8 . 6 . sin 45)

A = 16.97

### 11.4 Three-Dimensional Figures

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Describe the cross section formed by the intersection of the plane and the solid.

Question 18.

Answer:

The cross section is a rectangle.

Question 19.

Answer:

The cross-section is a square.

Question 20.

Answer:

the cross-section is a triangle.

### 11.5 Volumes of Prisms and Cylinders

Find the volume of the solid.

Question 21.

Answer:

Volume = lbh

V = 3.6 x 2.1 x 1.5 = 113.4 m³

Question 22.

Answer:

Volume = πr²h

V = π(2)² x 8 = 100.53 mm³

Question 23.

Answer:

Pentagon area = 6.88

Volume = Area x height

V = 6.88 x 4 = 27.52 yd³

### 11.6 Volumes of Pyramids

Find the volume of the pyramid.

Question 24.

Answer:

Volume V = Base area x height/3

Base Area = 9² = 81

V = 81 x 7/3 = 189 ft³

Question 25.

Answer:

Volume V = Base area x height/3

Base Area = 4 x 15 = 60

V = 60x 20/3 = 400 yd³

Question 26.

Answer:

Volume V = Base area x height/3

base area = 18 x 10 = 180

V = 180 x 5/3 = 300 m³

Question 27.

The volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.

Answer:

The side length of the square base is 3.46 in

Explanation:

The volume of a square pyramid is 60 cubic inches

V = 60

s²h/3 = 60

s² x 15/3 = 60

s² = 12

s = 3.46

Question 28.

The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid

Answer:

The volume of a square pyramid is 1024 cubic inches

s²h/3 = 1024

16²h = 3072

h = 12

### 11.7 Surface Areas and Volumes of Cones

Find the surface area and the volume of the cone.

Question 29.

Answer:

Surface area is 678.58 cm²

volume is 1017.87 cm³

Explanation:

Surface area of cone S = πr² + πrl

S = π x 9² + π x 9 x 15

S = 678.58

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 9² x 12)

V = 1017.87

Question 30.

Answer:

Surface area is 2513.27 cm²

volume is 8042.47 cm³

Explanation:

Surface area of cone S = πr² + πrl

S = π x 16² + π x 16 x 34

S = 2513.27

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 16² x 30)

V = 8042.47

Question 31.

Answer:

Surface area is 439.82 m²

volume is 562.102 m³

Explanation:

Surface area of cone S = πr² + πrl

S = π x 7² + π x 7 x 13

S = 439.82

h = √13² – 7² = 10.95

Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)

V = \(\frac { 1 }{ 3 } \)(π x 7² x 10.95)

V = 562.102

Question 32.

A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.

Answer:

The height of the cone = 15 cm.

Explanation:

r = 8

Volume V = 320π

\(\frac { 1 }{ 3 } \)(πr²h) = 320π

\(\frac { 1 }{ 3 } \)(π x 8² x h) = 320π

h = 15

### 11.8 Surface Areas and Volumes of Spheres

Find the surface area and the volume of the sphere.

Question 33.

Answer:

The surface area is 615.75 in², volume is 1436.75 in³

Explanation:

Surface area S = 4πr²

S = 4π x 7²

S = 615.75

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 7³

V = 1436.75

Question 34.

Answer:

The surface area is 907.92 ft², volume is 2572.44 ft³

Explanation:

d = 17

r = 8.5

Surface area S = 4πr²

S = 4π x 8.5²

S = 907.92

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 7³

V = 2572.44

Question 35.

Answer:

The surface area is 2827.43 ft², volume is 14137.16 ft³

Explanation:

C = 30π

2πr = 30π

r = 15

Surface area S = 4πr²

S = 4π x 15²

S = 2827.43

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 15³

V = 14137.16

Question 36.

The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.

Answer:

The surface area and the volume of Mercury is 23814400π, 19369045330π

Explanation:

d = 4880

r = 2440

Surface area S = 4πr²

S = 4π x 2440² = 23814400π

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 2440³

V = 19369045330π

Question 37.

A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.

Answer:

Volume of the composite solid = 272.52

Explanation:

Volume of cube = a³

= 6³ = 216

Volume of hemisphere = \(\frac { 4 }{ 6 } \)πr³

= \(\frac { 4 }{ 6 } \)π x 3³ = 18π

Volume of the composite solid = 216 + 18π = 272.52

### Circumference, Area, and Volume Test

Find the volume of the solid.

Question 1.

Answer:

Volume = 2577.29 m³

Explanation:

Volume = \(\frac { 3√3 }{ 2 } \)a²h

= \(\frac { 3√3 }{ 2 } \) x 8² x 15.5

= 2577.29

Question 2.

Answer:

Volume is 17.157 ft³

Explanation:

d = 3.2

r = 1.6

Volume V = \(\frac { 4 }{ 3 } \)πr³

V = \(\frac { 4 }{ 3 } \)π x 1.6³

V = 17.157

Question 3.

Answer:

Volume of sloid = 402.11 m³

Explanation:

Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

= \(\frac { 1 }{ 3 } \)π x 4² x 3

= 50.26

Volume of cylinder = πr²h

= π x 4² x 6 = 301.59

Volume of sloid = 2(50.26) + 301.59 = 402.11

Question 4.

Answer:

Volume of solid = 106.66

Explanation:

Volume of rectangular box = 5 x 2 x 8 = 80

Volume of pyramid = 80/3 = 26.66

Volume of solid = 80 + 26.66 = 106.66

Find the indicated measure.

Question 5.

circumference of ⊙F

Answer:

circumference of ⊙F is 109.71 in

Explanation:

64 = \(\frac { 210 }{ 360 } \) • C

C = 109.7

Question 6.

m\(\widehat{G H}\)

Answer:

m\(\widehat{G H}\) = 74.27

Explanation:

35 = \(\frac { x }{ 360 } \) • 2π x 27

x = 74.27

Question 7.

area of shaded sector

Answer:

Area is 142.41 in²

Explanation:

Area = \(\frac { 360 – 105 }{ 360 } \) • π x 8²

Area = 142.41

Question 8.

Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.

Answer:

The radius of the cylinder is 3

The height of the cylinder is 6.

The radius of the hemisphere is 3.

Question 9.

Find the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.

Answer:

The surface area is 486.7 sq ft

Explanation:

l² = r² + h²

l² = 5² + 12²

l = 13

Surface area S = πr² + 2πrl

S = π x 5² + 2π x 5 x 13

S = 486.7

Question 10.

You have a funnel with the dimensions shown.

a. Find the approximate volume of the funnel.

Answer:

Volume = \(\frac { 1 }{ 3 } \)πr²h

= \(\frac { 1 }{ 3 } \)π x 6² x 10

= 376.99

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm^{3})

Answer:

T = 376.8 ml/45 ml per sec

T = 8.373 sec

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second?

Answer:

V = 1/3 πr²h

= 1/3 (3.14 × 10² × 6)

= 1/3(1884)

= 628 cu. cm

T = 628 ml/45 ml per sec

T = 13.95 sec

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.

Answer:

- In cone type shaped object if the radius is large then the volume of the cone increases.
- In the b part of the funnel, the radius is smaller than the height and in the c part, the radius is larger than the height of the cone.
- V = 1/3 πr²h
- It means if radius increase volume also increases with the same rate. So, the time is taken by the c part (13.95 sec) is larger than the b part.

Question 11.

A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.

Answer:

The height of the bottle is 11.3 cm

Explanation:

Volume of cylinder = 500

πr²h = 500

π(3.75)²h = 500

h = 11.3 cm

Question 12.

Find the area of a dodecagon (12 sides) with a side length of 9 inches.

Answer:

Area is 237.31

Explanation:

Area = \(\frac { 1 }{ 4 } \)πa²cot(π/n)

= \(\frac { 1 }{ 4 } \)π x 9² x cot(π/12)

= 237.31

Question 13.

In general, a cardboard fan with a greater area does a better job of moving air and cooling you. The fan shown is a sector of a cardboard circle. Another fan has a radius of 6 centimeters and an intercepted are of 150°. Which fan does a better job of cooling you?

Answer:

Area of sector for fan of radius 9 cm = 120/360 × 3.14 × 9 × 9

= 1/3 × 254.57

= 84.85 sq. cm

Area of sector of fan of radius 6 cm = 150/360 × 3.14 × 6 × 6 = 5/12 × 113.14 = 47.14 sq. cm

### Circumference, Area, and Volume Cumulative Assessment

Question 1.

Identify the shape of the cross section formed by the intersection of the plane and the

solid.

a.

Answer:

The cross-section is a trapezoid.

b.

Answer:

The cross-section is a pentagon.

c.

Answer:

The cross-section is a rectangle.

Question 2.

In the diagram, is tangent to ⊙P at Q and \(\overline{P Q}\) is a radius of ⊙P? What must be true about and \(\overline{P Q}\)? Select all that apply.

Answer:

PQ is perpendicular to RS.

Question 3.

A crayon can be approximated by a composite solid made from a cylinder and a cone.

A crayon box is a rectangular prism. The dimensions of a crayon and a crayon box containing 24 crayons are shown.

a. Find the volume of a crayon.

Answer:

The volume of a crayon = πr²h + \(\frac { 1 }{ 3 } \)πr²h

= π x 4.25² x 80 + \(\frac { 1 }{ 3 } \)π x 3.25² x 10

= 4539.6 + 110.61

= 4650.21 mm³

b. Find the amount of space within the crayon box not taken up by the crayons.

Answer:

Volume of box = 94 x 28 x 71 = 186872

The volume of a crayon = 4650.21

Remaining space = 186872 – 24 x 4650.21

= 75266.96

Question 4.

What is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \(\frac{1}{2}\)y = – 1?

(A) y = – 2x + 9

(B) y = 2x + 1

(C) y = \(\frac{1}{2}\)x + 4

(D) y = –\(\frac{1}{2}\)x + 6

Answer:

(A) y = – 2x + 9

Explanation:

x + \(\frac{1}{2}\)y = – 1

y = -2 – 2x

The slope of the line is -2

The euation of line is y – 5 = -2(x – 2)

y – 5 = -2x + 4

y = -2x + 9

Question 5.

The top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion?

(A) 22,019.63 ft^{3}

(B) 172,006.91 ft^{3}

(C) 66,058.88 ft^{3}

(D) 207,530.08 ft^{3}

Answer:

(A) 22,019.63 ft^{3}

Explanation:

Volume = a²\(\frac { h }{ 3 } \)

= 34.5² x \(\frac { 55.5 }{ 3 } \)

= 22019.62

Question 6.

Prove or disprove that the point (1, √3 ) lies on the circle centered at the origin and containing the point (0, 2).

Answer:

We consider the circle centered at the origin and containing the point (0, 2).

Therefore, we canconclude that rdaius is 2 and points be (0, 0), (1, √3)

distance = √(1 – 0)² + (√3 – 0)² = 2

As radius and distance are same. The point B(1, √3) lies on the circle.

Question 7.

Your friend claims that the house shown can be described as a composite solid made from a rectangular prism and a triangular prism. Do you support your friend’s claim? Explain your reasoning.

Answer:

Yes

Question 8.

The diagram shows a square pyramid and a cone. Both solids have the same height, h, and the base of the cone has radius r. According to Cavalieri’s Principle, the solids will have the same volume if the square base has sides of length ______ .

Answer:

2r/√2

Explanation:

Volume of square pyramid = a²\(\frac { h }{ 3 } \)

Square diagonal = √2a

radius = √2a/2

a = 2r/√2

Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

Question 9.

About 19,400 people live in a region with a 5-mile radius. Find the population density in people per square mile.

Answer:

The number of people per square mile is 247

Explanation:

S = πr²

= π x 5² = 78.5

Number of people per square mile = 19400/78.5 = 247