12-1 Circles and Circumference · Find the circumference of each circle. Round to the nearest tenth. 7. SOLUTION: The circumference is about 22.0 meters. 8. SOLUTION: The circumference

Find the circumference of each circle. Round to the nearest tenth.

7.

SOLUTION:

The circumference is about 22.0 meters.

8.

SOLUTION:

The circumference is about 6.3 centimeters.

9.

SOLUTION:

The circumference is about 62.8 feet.

10.

SOLUTION:

The circumference is about 50.3 inches.

11.

SOLUTION:


12.

SOLUTION:

The circumference is about 39.0 kilometers.

13. diameter = 9.4 meters SOLUTION:


14. radius = 13.7 millimeters SOLUTION:

The circumference is about 86.1 millimeters.

15. radius = feet

SOLUTION:


16. diameter = yards

SOLUTION:

The circumference is about 45.9 yards.

17. Compare the circumferences of the two sand dollars shown.

SOLUTION:

The diameter of the small sand dollar is 1 inches.

The circumference of the small sand dollar is about 3.9 inches.

The diameter of the large sand dollar is 2 inches.

The circumference of the large sand dollar is about 7.9 inches. Since 3.9 is one half of 7.9, the circumference of the small sand dollar is half the circumference of the large sand dollar.

18. The circumference of the Moon is about 6790 miles. What is the distance to the center of the Moon in kilometers? Round to the nearest kilometer. (Hint: 1 mile ≈ 1.6 kilometers)

SOLUTION: The distance to the center of the Moon is the radius of the Moon. Substitute 6790 for C in the equation C = 2πr.

So, the radius of the moon is about 1080.7 miles. Since 1 mile ≈ 1.6 kilometers, the moon is about 1080.7 • 1.6 or 1729 km.

19. The tower at Philadelphia City Hall contains four clocks that have a radius of about 3.96 meters. Find how far the minute hand travels after each number of rotations around the clock face. Round to the nearest hundredth. a. 2 rotations

b. rotation

c. rotations

SOLUTION: First find the circumference of a clock.

So, if the minute hand makes 1 rotation it travels about 24.88 meters. a. If the minute hand travels 2 rotations it travels

or 49.76 meters.

b. If the minute hand travels rotation it travels

or 12.44 meters.

c. If the minute hand travels rotations it travels

or 143.07 meters.

20. The world’s largest carousel in Spring Green, Wisconsin, has a diameter of 80 feet. How far does a rider travel on an outside horse after 10 revolutions? Round to the nearest foot.

SOLUTION: First find the circumference of the carousel.

So, a rider will travel 251.33 feet in 1 revolution. If they make 10 revolutions, they will travel 10 • 251.33 or 2513 feet.

21. A circular fountain at a park has a radius of 4 feet. The mayor wants to build a fountain that is quadruplethe size of the current fountain. Find the circumference of the new fountain. Round to the nearest tenth.

SOLUTION: First find the circumference of the current fountain.

The circumference of the current fountain is 25.13 feet. The circumference of the new fountain is 4 • 25.13 or 100.5 feet.

30. A plate has a radius of 5 inches. Which equation could be used to find the circumference of the plate in inches?

A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:

The equation C = 10π could be used to find the circumference of the plate. Choice C is correct.

31. A bicycle tire has a diameter of 24 inches. Find the circumference of the tire to the nearest tenth of an inch.

F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:

The circumference is about 75.4 inches. Choice F is correct.

32. A planter has a circumference of 37.6 inches. Whichmeasure is closest to the diameter of the planter?

A 10 in.B 11 in.C 12 in.D 14 in.

SOLUTION: Substitute 37.6 for C in the equation C = πd.

The diameter of the circle is about 12 inches. Choice C is correct.

33. Short Response The circle shown below has a diameter of 15 feet.

What is the circumference to the nearest foot?

SOLUTION:

The circumference is about 47 feet.


7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:


eSolutions Manual - Powered by Cognero Page 1

12-1 Circles and Circumference


7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:



7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:





7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:



7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:





7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:



7.

SOLUTION:


8.

SOLUTION:


9.

SOLUTION:


10.

SOLUTION:


11.

SOLUTION:


12.

SOLUTION:






15. radius = feet

SOLUTION:



SOLUTION:



SOLUTION:









b. rotation

c. rotations



or 49.76 meters.


or 12.44 meters.


or 143.07 meters.








A C = 2(10π)B C = 5πC C = 10πD C = 2π

SOLUTION:



F 75.4 in.G 83.5 in.H 87.9 in.J 91.5 in.

SOLUTION:



A 10 in.B 11 in.C 12 in.D 14 in.





SOLUTION:




Find the area of each circle. Round to the nearest tenth.

9.

SOLUTION:

The area is about 706.9 square inches.

10.

SOLUTION: The radius is half the diameter, so 10 ÷ 2 = 5.

The area is about 78.5 square miles.

11.

SOLUTION: The radius is half the diameter, so 17 ÷ 2 = 8.5.

The area is about 227.0 square feet.

12.

SOLUTION:

The area is about 66.5 square centimeters.

13.

SOLUTION: The radius is half the diameter, so 20.3 ÷ 2 = 10.15.

The area is about 323.7 square meters.

14. radius = 9.6 feet SOLUTION:


15. diameter = 24.8 meters SOLUTION: The radius is half the diameter, so 24.8 ÷ 2 = 12.4.



SOLUTION:

The radius is half the diameter, so 11 ÷ 2 = 5 or

5.75.

The area is about 103.9 square yards.

17. radius = miles

SOLUTION:


18. Each shelf of a shelving unit is a quarter circle with aradius of 32 centimeters. What is the area of each shelf? Round to the nearest tenth.

SOLUTION: The area of the quarter circle is one-fourth the area of a circle.

The area of each shelf is about 804.2 square centimeters.

19. Lauren has a sprinkler positioned in her lawn that directs a 12-foot spray in a circular pattern. About how much of the lawn does the sprinkler water if there is a rectangular flower bed 3 feet by 6 feet thatis also in the path of the spray?

SOLUTION: Find the areas of the circle and of the rectangle. Then subtract the area of the rectangle from the areaof the circle.

So, the sprinkler waters about 452.4 – 18 or about 434.4 square feet of the lawn.

20. What is the area of the CD shown below? Round to the nearest tenth.

SOLUTION: Find the areas of the larger circle and of the smaller circle. Then subtract the area of the smaller circle from the area of the larger circle. For the larger circle the radius is half the diameter so12 ÷ 2 = 6.

For the smaller circle the radius is half the diameter so 1.5 ÷ 2 = 0.75.

So, the CD is about 113.1 – 1.8 or about 111.3 square centimeters.

21. The trunk of the General Sherman Tree in Sequoia National Park has a circumference of 102.6 feet. If the tree were cut down at the base, what would be the area of the cross section?

SOLUTION: First use the formula for circumference to find the radius of the tree trunk. Substitute 102.6 for circumference.

The radius of the trunk is about 16.3293 feet.

The area of the cross section would be about 837.7 square feet.

34. Find the area of a circle with a diameter of 22 millimeters. Round to the nearest tenth.

A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2


The area is about 380.1 square millimeters. Choice Ais correct.

35. A sprinkler is set to cover the area shown. Find the area of the grass being watered if the sprinkler reaches a distance of 10 feet.

F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2

SOLUTION: First find the area of the whole circle, then subtract one fourth of the area to find the area of the grass being watered.

So the area of the grass being watered is 314.16 – 314.16 ÷ 4 or 235.6 square feet. Choice H is correct.

36. The Blackwells have a circular pool with a radius of 10 feet. They want to install a 3-foot sidewalk aroundthe pool. What will be the area of the walkway?

A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2

SOLUTION: To find the area of the walkway, find the area of the whole circle and then subtract the area of the inner circle. The whole circle has a radius of 13 feet.

The inner circle has a radius of 10 feet.

The area of the walkway is 530.93 – 314.16 or 216.8square feet. Choice A is correct.

37. Short Response The area of a circle is 327.6 square centimeters. a. Write an algebraic expression in terms of A that could be used to find the radius of the circle. b. Find the radius to the nearest tenth.

SOLUTION: a.

b.

The radius of the circle is about 10.2 centimeters.


9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.



12-2 Area of Circles


9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.



9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.





9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.



9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.





9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.



9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.





9.

SOLUTION:


10.



11.



12.

SOLUTION:


13.








SOLUTION:


5.75.


17. radius = miles

SOLUTION:

















A 380.1 mm2

B 319.5 mm2

C 189.9 mm2

D 69.1 mm2




F 47.1 ft2

G 157.1 ft2

H 235.6 ft2

J 314.2 ft2




A 216.8 ft2

B 285.9 ft2

C 314.2 ft2

D 442.2 ft2





SOLUTION: a.

b.




Find the area of each figure. Round to the nearest tenth, if necessary.

7.

SOLUTION: Separate the figure into a rectangle and a triangle. Find the sum of the areas of the figures.

The base of the triangle is 22 – 4 or 18 inches. The height of the triangle is 18 – 6 or 12 inches.

So, the area of the composite figure is 132 + 108 or 240 square inches.

8.

SOLUTION: Separate the figure into a square and a trapezoid. Find the sum of the areas of the figures.

So, the area of the composite figure is 25 + 14 or 39 square feet.

9.

SOLUTION: Separate the figure into a rectangle, a semicircle, anda triangle. Find the sum of the areas of the figures.

The radius of the semicircle is half the diameter, so 2.4 ÷ 2 = 1.2.

So, the area of the composite figure is 3.84 + 2.3 + 1.2 or about 7.3 square millimeters.

10.

SOLUTION: Separate the figure into a trapezoid and a semicircle. Find the sum of the areas of the figures.

The radius of the semicircle is half the diameter, so 9÷ 2 = 4.5.

So, the area of the composite figure is 121.5 + 31.8 or about 153.3 square inches.

11.

SOLUTION: Separate the figure into a square and four semicircles. Find the sum of the areas of the figures. The diameter of the circles and the sides of the square are each equal to 20 ÷ 2 or 10 feet.

The radius of the semicircle is half the diameter, so 10 ÷ 2 = 5.

There are 4 semicircles, so the area of the semicircles is 4 • 39.27 or about 157.1 square feet. So, the area of the composite figure is 100 + 157.1 orabout 257.1 square feet.

12. Refer to the swimming pool shown.

a. What is the area of the pool’s floor? b. A pool cover costs $1.70 per square foot. How much would a cover for the pool cost?

SOLUTION: a. Separate the figure into two trapezoids. Find the sum of the areas of the figures.

So, the area of the pool’s floor is 175 + 213.75 or 388.75 square feet. b. To determine the cost of the pool cover, find the product of the pool’s area and the cost per square foot. 388.75 • 1.7 ≈ 660.88 The pool cover will cost about $660.88.

13. Financial Literacy Mr. Reyes wants to carpet his family room.

a. What is the area of the space to be carpeted? b. If carpet costs $2.25 per square foot, how much would it cost to carpet Mr. Reyes’ family room if there is no leftover carpet?

SOLUTION: a. Separate the figure into a rectangle and a trapezoid. Find the sum of the areas of the figures.

The base of the trapezoid is 23 – 7 or 16 feet. The height of the trapezoid is 18 – 12 or 6 feet.

So, the area of the space to be carpeted is 276 + 78 or 354 square feet. b. To determine the cost of the carpet, find the product of the area of the room and the cost per square foot. 354 • 2.25 = 796.5 The carpet will cost $796.50.

14. What is the area of a figure that is formed using a rectangle 10 feet long and 7 feet wide and a semicircle with a diameter of 7 feet?

SOLUTION: Find the sum of the areas of the figures.


So, the area of the composite figure is 70 + 19.2 or about 89.2 square feet.

15. A figure is formed using a semicircle and a triangle that has a base of 8 inches and a height of 12 inches.Find the area of each figure to the nearest tenth. a. The diameter of the semicircle equals the base of the triangle. b. The diameter of the semicircle equals the height ofthe triangle.

SOLUTION: a. Separate the figure into a semicircle and a triangle.Find the sum of the areas of the figures. The radius of the semicircle is half the diameter, so 8÷ 2 = 4.

So, the area of the composite figure is 25.1 + 48 or about 73.1 square inches. b. Separate the figure into a semicircle and a triangle. Find the sum of the areas of the figures. The radius of the semicircle is half the diameter, so 12 ÷ 2 = 6.

So, the area of the composite figure is 56.5 + 48 or about 104.5 square inches.

27. Short Response In the diagram, a patio that is 4 feet wide surrounds a swimming pool. What is the area of the patio in square feet? Round to the nearesttenth.

SOLUTION: To find the area of the patio, subtract the area of the pool from the total area of the composite figure. Separate the figure into a semicircle and a rectangle. Find the sum of the areas of the figures. The radius of the semicircle is half the diameter, so 24 ÷ 2 = 12.

The total area is 226.2 + 720 or about 946.2 square feet.

The area of the pool is 100.5 + 416 or about 516.5 square feet. So, the area of the patio is 946.2 – 516.5or 429.7 square feet.

28. Find the area of the shaded region. Use 3.14 for π. Round to the nearest hundredth.

A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2

SOLUTION: To find the area of the shaded region, subtract the area of the square from the area of the circle. The radius of the circle is half the diameter, so 19.8 ÷ 2 = 9.9.

So, the area of the shaded region is 307.75 – 196 or about 111.75 square feet. Choice C is correct.

29. The Lin family is buying a cover for their swimming pool shown. The cover costs $3.19 per square foot. How much will the cover cost?

F $219.27G $258.54H $699.47J $824.74

SOLUTION: Separate the figure into a rectangle and two semicircles. Find the sum of the areas of the figures.

If the two semicircles are combined, they form a complete circle. The radius of the circle is half the diameter, so 10 ÷ 2 = 5.

So, the area of the composite figure is 180 + 78.54 orabout 258.54 square feet. To determine the cost of the pool cover, find the product of the area of the pool and the cost per square foot. 258.54 • 3.19 ≈ 824.74 The pool cover will cost about $824.74. Choice J is correct.

30. Find the difference between the areas of the two bedrooms (not including the closet).

A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:

The area of bedroom 2 is equal to the difference of the total area and the area of the closet.

The difference between the areas of the two bedrooms is 150 – 110 or 40 square feet. Choice B iscorrect.


7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




12-3 Area of Composite Figures


7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:






7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:






7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:






7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:






7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:




7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:






7.




8.



9.




10.




11.


























A 392.04 ft2

B 503.75 ft2

C 111.75 ft2

D 258.17 ft2




F $219.27G $258.54H $699.47J $824.74





A 27 ft2

B 40 ft2

C 58 ft2

D 168 ft2

SOLUTION:





Documents

12-1 Circles and Circumference · Find the circumference of each circle. Round to the nearest tenth. 7. SOLUTION: The circumference is about 22.0 meters. 8. SOLUTION: The circumference