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Find the surface area of each solid. Round to thenearest tenth if necessary.

1.

SOLUTION:

The surface area S of a regularpyramid is , where L is the lateral area and B is the area of the base.

Therefore, the surface area of the pyramid is 640

cm2.

ANSWER:

640 cm2

2.

SOLUTION: Surface area = 2(10 m)(11 m) + 2(10 m)(15 m) +2(11 m)(15 m) = 850 m²

ANSWER: 850 m²

3.

SOLUTION:

The radius is 7.5 cm and the height is 15 cm.

Use the Pythagorean Theorem to find the slantheight.

Find the surface area.

ANSWER:

≈ 571.9 cm2

4.

SOLUTION:

The segment joining the points where the slant heightand height intersect the base is the apothem.

eSolutions Manual - Powered by Cognero Page 1

10-6 Surface Area

Use the Pythagorean Theorem to find the length theapothem of the base.

A central angle of the regular hexagon is ,so the angle formed in the triangle below is 30.

Use a trigonometric ratio to find the length of a sideof the hexagon s.

Find the surface area of the pyramid.

Therefore, the surface area of the pyramid is about

332.6 m2.

ANSWER:

≈ 332.6 m2

5.

SOLUTION:

The base of the prism is a right triangle with the legs8 ft and 6 ft long. Use the Pythagorean Theorem tofind the length of the hypotenuse of the base.


ANSWER:

336 ft2


10-6 Surface Area

6.

SOLUTION:

ANSWER:

≈ 2063.6 cm2

7. CARS Evan is buying new tire rims that are 14inches in diameter and 6 inches wide. Determine thesurface area of each rim. Round to the nearest tenth.

SOLUTION:

ANSWER:

571.8 in2

8. PATIO STONES A patio stone has a rectangularbase that is 3 inches by 8 inches and a height of 4inches. What is the surface area of the stone?

SOLUTION:

ANSWER:

9. ROOFING A pyramid shaped roof has a squarebase that is 30 feet wide and a slant height of 14 feet.How much roofing material is needed to cover theroof?

SOLUTION: ROOFING A pyramid shaped roof has a squarebase that is 30 feet wide and a slant height of 14 feet.How much roofing material is needed to cover theroof? The amount of roofing material needed to cover theroof is the lateral area of the pyramid. L = (0.5)Pl = (0.5)(4)(30 ft)(14 ft) = 840 ft²

ANSWER: 840 ft²


10-6 Surface Area

10. TANKS A storage tank is shown at the right.Round answers to the nearest tenth.

a. Find the lateral area of the cylinder. b. Find the lateral area of the cone. c. Find the total lateral area of the tank.

SOLUTION:

a. Find the lateral area of a cylinder with a radius of feet and a height of 11 feet.

b. Find the lateral area of a cone with a radius of 7feet and a slant height of 9 feet.

c. Find the total lateral area of the tank by addingthe lateral area of the cylinder and the lateral area ofthe cone.

ANSWER:

a. 483.8 ft²b. 197.9 ft²c. 681.7 ft²

Find the surface area of each solid. Round to thenearest tenth if necessary.

11.

SOLUTION:

Find the length of the third side of the triangle.

Now find the surface area.

ANSWER:

36 ft2


10-6 Surface Area

12.

SOLUTION:

The radius of the cone is and the heightis 18 cm. Use the Pythagorean Theorem to find theslant height .

Find the surface areas of the cone.

Therefore, the surface area of the cone is about

470.7 cm2.

ANSWER:

≈ 470.7 cm2

13.

SOLUTION:

The diameter of the base is 3.6 cm and the height ofthe cylinder is 1.1 cm.

The total surface area of the prism is the sum of theareas of the bases and the lateral surface area.

ANSWER:

≈ 32.8 cm2

14.

SOLUTION:

We need to find the area of the triangle to determinethe area of the bases. Use the Pythagorean Theoremto find the height of the triangles.


10-6 Surface Area

Use to calculate the surface area.

ANSWER:

151.9 m2

15.

SOLUTION:

The base of the pyramid is a square with side lengthof 8 feet and the height is 10 feet. Use thePythagorean Theorem to find the slant height of thepyramid.



236.3 ft2.

ANSWER:

≈ 236.3 ft2

16.

SOLUTION:

Use the right triangle formed by the slant height of 7,the height of 5, and the apothem and the Pythagorean


10-6 Surface Area

Theorem to find the length of the apothem of thebase.

The base of the pyramid is an equilateral triangle.

The measure of each central angle is , so

the angle formed in the triangle below is 60°.

Use a trigonometric ratio to find the length s of eachside of the triangle.

Use the formulas for regular polygons to find theperimeter and area of the base.



302.9 cm2.

ANSWER:

≈ 302.9 cm2

17. A cone has a diameter of 3.4 centimeters, and a slantheight of 6.5 centimeters.

SOLUTION:

The radius of the cone is .


Therefore, the surface area of the cone is about 43.8cm2.

ANSWER:

≈ 43.8 cm2


10-6 Surface Area

18. A rectangular prism has = 25 centimeters, w = 18centimeters, and h = 12 centimeters.

SOLUTION:

Note that the surface area of the solid is the same nomatter which face is the base.

ANSWER:

1932 cm2

19. A regular hexagonal pyramid has a base edge of 6millimeters and a slant height of 9 millimeters.

SOLUTION:

The base of the pyramid is a regular hexagon. Theperimeter of the hexagon is P = 6 × 6 or 36 mm.

A central angle of the hexagon is , so theangle formed in the triangle below is 30°.

Use a trigonometric ratio to find the apothem a andthen find the area of the base.

So, the area of the base B is about 93.5 mm2. Find the surface area of the pyramid.


255.5 mm2.

ANSWER:

≈ 255.5 mm2


10-6 Surface Area

20. A triangular prism has h = 6 inches and a righttriangle base with legs 9 inches and 12 inches.

SOLUTION:

Find the other side of the triangular base.

Now you can find the urface area.

ANSWER:

324 in2

21. A cylinder has a diameter of 8 inches and a height of6.2 inches.

SOLUTION:

ANSWER:

≈ 256.4 in.2

22. A square pyramid has an altitude of 12 inches and aslant height of 18 inches.

SOLUTION:

The base of the pyramid is square. Use thePythagorean Theorem to find the length the apothemof the base.

A central angle of the square is , so theangle formed in the triangle below is 45°.

Use a trigonometric ratio to find the length of eachside of the square.

Find the perimeter and area of the base.



10-6 Surface Area


1686.0 in2.

ANSWER:

≈ 1686.0 in2

23. A cylinder has a radius of 3 millimeters and a heightof 15 millimeters.

SOLUTION:

ANSWER:

≈ 339.3 mm2

24. A cone has an altitude of 5 feet, and a slant height of

feet.

SOLUTION:

The altitude of the cone is 5 feet and the slant height

is or 9.5 feet. Use the Pythagorean Theorem to

find the radius.

Find the surface area of the cone.

Therefore, the surface area of the cone is about446.1 ft2.

ANSWER:

≈ 446.1 ft2

25. Find the lateral area of the tent to the nearest tenth.

SOLUTION:

The tent is a combination of a cylinder and a cone.


10-6 Surface Area

The cone has a radius of 5 feet and a height of 6feet. Use the Pythagorean Theorem to find the slantheight of the cone.

The cylinder has a radius of 5 feet and a height of 12– 6 or 6 feet. Find the sum of the lateral areas of the cone andcylinder to find the lateral area of the tent.

Therefore, the lateral area of the tent is about 311.2

ft2.

ANSWER:

311.2 ft2

26. Find the lateral area of the dog house with a 12 inchsquare cut out of one face for the door.

SOLUTION: Start by finding the lateral area of the walls(rectangular prism) and then subtracting the area ofthe door:

The lateral area of the walls minus the door is .

Next, find the lateral area of the roof (squarepyramid):

The total lateral area of the dog house is thecombination of the walls ( minus the door) and theroof. Combine the lateral areas of each to get thetotal lateral area:

ANSWER: 6465.1 in²


10-6 Surface Area

Find the surface area of each composite solid.Round to the nearest tenth if necessary.

27.

SOLUTION:

The solid is a combination of a rectangular prism anda cylinder. The base of the rectangular prism is 6 inby 4 in and the radius of the cylinder is 3 in. Theheight of the solid is 15 in. Rectangular prism:

The surface area of five faces of the rectangular

prism is 258 cm2. Half-cylinder:

The surface area of the half-cylinder is 54π cm2. The total surface area is 258 + 54π = 427.6.

ANSWER:

427.6 in2

28.

SOLUTION:

The solid is a combination of a cube and a cylinder.The length of each side of the cube is 12 cm and theradius of the cylinder is 6 cm. The height of the solidis 12 cm. Rectangular prism:

The surface area of five faces of the rectangular

prism is 720 cm2. Half-cylinder:

The surface area of the half-cylinder is 108π cm2 and

the total surface area is about 1059.3 cm2.

ANSWER:

1059.3 cm2


10-6 Surface Area

29. MOUNTAINS A conical mountain has a radius of1.6 kilometers and a height of 0.5 kilometer. What isthe lateral area of the mountain?

SOLUTION:

The radius of the conical mountain is 1.6 kilometersand the height is 0.5 kilometers. Use the PythagoreanTheorem to find the slant height.

Find the lateral area L of the conical mountain.

Therefore, the lateral area is about 8.4 km2.

ANSWER:

8.4 km2

30. AQUARIUMS The Tower Aquarium in HenleyBeach, Australia, is the world’s largest cylindricalaquarium. It reaches a height of over 40 meters andis 36 meters in diameter. Visitors ascend through acolumn of water as they ride a split-level glass lift upseven floors, through the center of the aquarium.What is the approximate lateral area of the outsideof the aquarium?

SOLUTION:

Since the diameter of the cylinder is 36 m, the radiusis .

ANSWER:


10-6 Surface Area

31. HISTORY Archaeologists recently discovered a1500-year-old pyramid in Mexico City. The squarepyramid measures 165 yards on each side and oncestood 20 yards tall. What was the original lateral areaof the pyramid?

SOLUTION:

The pyramid has a square base with sides havinglengths of 165 yards and a height of 20 yards. Usethe Pythagorean Theorem to find the slant height.

Find the lateral area L of a regular pyramid.

Therefore, the lateral area of the pyramid is about28,013.6 yd2.

ANSWER:

28,013.6 yd2

32. MONUMENTS The monolith mysteriouslyappeared overnight at Seattle, Washington’sManguson Park. It is a hollow rectangular prism 9feet tall, 4 feet wide, and 1 foot thick.a. Find the area in square feet of the structure’ssurfaces that lie above the ground.b. Use dimensional analysis to find the area in squareyards.

SOLUTION:

a. The area of surfaces that lie above the ground isthe sum of the area of the upper base and the lateralsurface area.

b.

ANSWER:

a. 94 ft2

b. 10.4 yd2

33. TEPEES The dimensions of two canvas tepees areshown in the table at the right. Including the floors,approximately how much more canvas is used tomake Tepee B than Tepee A?

SOLUTION:

The tepees are in the shape of a right cone. To findthe amount of canvas used for each tepee, we need


10-6 Surface Area

to find its lateral area. The radius of Tepee A is

or 7 feet and the radius of Tepee B is or 10 feet

Tepee A Tepee B Use the Pythagorean Theorem to find the slant height of each tepee and then find the surface area for

each cone.

To find how much more canvas is used to makeTepee B than Tepee A, subtract the surface areas.

Tepee B will use about 380.1 ft2 more canvas thanTepee A.

ANSWER:

about 380.1 ft2

34. DESIGN A mailer needs to hold a poster that isalmost 38 inches long and has a maximum rolleddiameter of 6 inches.a. Design a mailer that is a triangular prism. Sketchthe mailer and its net. b. Suppose you want to minimize the surface area ofthe mailer. What would be the dimensions of themailer and its surface area?

SOLUTION: a. A triangular prism should consist of two trianglesand three rectangles. They should be connected sothat, when folded together they form a prism.

b. In order to minimize the surface area of thetriangular prism, the triangles should be equilateral,and the side lengths of the rectangles should coincidethe the length of the base of the triangle and thelength of the poster. The surface area will then be

. Use trigonometry to find the area of the triangles.The diameter of the poster has a maximum of 6 in.


10-6 Surface Area

which corresponds to a radius of 3 in.

For the base we have:

For the height we have:

Now calculate the area:

The area of the rectangles will be the product of thearea of the base of the triangle with the length of theposter.

The total surface area can be calculated:

ANSWER: a. Sample answer:

b. side lengths of triangular bases, about 10.39

in.each; height, 38 in.; 1278 in2

35. MULTI-STEP Hector is designing a glassgreenhouse for a city park. He has a 40-foot by 20-foot rectangular plot available. He wants the roof tobe a triangular prism in which the center of the roof is4 feet higher than the edges. The glass costs $25 persquare foot, and Hector cannot spend more than$60,000 on glass.a. What is the maximum height that Hector shouldmake the edge of the roof? b. Describe your solution process. c. What assumptions did you make?

SOLUTION: a-b. Start by sketching a figure with the giveninformation, with g representing the height of theroof.


10-6 Surface Area

Find the sum of the surface areas of each individualsection. The rectangular section of the front and backis 2 × 20 × g or 40g ft². The sides cover 2 × 40 × gor 80g ft². The triangular tops of the front and backof the greenhouse cover 2(0.5)(4)(20) or 80 ft². The

slant of the roof is .Thus, the roof covers 2(40)(10.77) or 861 ft². The total surface area is 861 + 80 + 120g ft². Hector can use up to 60,000 ÷ 25 or 2400 ft². Therefore, g is approximately 12.1. Rounding down,we get a height of 12 ft. c. Hector used the entire available plot. There was noglass used for the base. The entrance was made ofglass. The top of the roof ran along the 40 ft length ofthe greenhouse.

ANSWER: a. about 12 ftb. First, find the sum of the surface areas of eachindividual section. The rectangular section of the frontand back is 2 × 20 × g or 40g ft². The sides cover 2 ×40 × g or 80g ft². The triangular tops of the front andback of the greenhouse cover 2(0.5)(4)(20) or 80 ft².The slant of the roof is . Thus, the roofcovers 2(40)(10.77) or 861 ft². The total surface areais 861 + 80 + 120g ft². Hector can use up to 60,000 ÷25 or 2400 ft². Therefore, g is approximately 12.1.Rounding down, we get a height of 12 ft.c. Hector used the entire available plot. There was noglass used for the base. The entrance was made ofglass. The top of the roof ran along the 40 ft length ofthe greenhouse.

36. PETS A frustum is the part of a solid that remains

after the top portion has been cut by a plane parallelto the base. The ferret tent shown is a frustum of aregular pyramid.a. Describe the faces of the solid.b. Find the surface area of the frustum formed by thetent.c. Another pet tent is made by cutting the top half offof a pyramid with a height of 12 centimeters, slantheight of 20 centimeters and square base with sidelengths of 32 centimeters. Find the surface area ofthe frustum.

SOLUTION:

a. The two bases are squares and the 4 lateral facesare trapezoids.b. Each lateral face is a trapezoid with the bases 6 in.and 17 in. and height 15 in. The area A of a trapezoidwith bases b1, b2 and the height h is given by the

formula

The lateral area of the solid is

The bases are squares of sides 6 in. and 17 in.respectively. Therefore, the surface area is

c. When the top half of a pyramid with a height of 12cm, slant height of 20 cm and square base with sidelengths of 32 cm is cut, the height of the frustum willbe 6 cm, the slant height 10 cm and the length ofeach side of the upper base will be 16 cm.


10-6 Surface Area

The total surface area of the frustum will be

ANSWER:

a. 4 trapezoids, 2 squaresb. 1015 in2

c. 2240 cm2

37. The three-dimensional box needs to have a clearcoating painted on all six faces. What is theapproximate surface area of the box?

SOLUTION:

This composite solid can be divided into a rectangularprism 13 cm by 21 cm by 28 cm and a triangularprism that has a right triangle with legs of 7 cm and21 cm as the base and a height of 28 cm. Thesurface area of the solid is the sum of the surfaceareas of each prism without the area of the 21 cm by28 cm rectangular face at which they are joined.

.Rectangular prism:

So, the surface area of five faces of the rectangularprism is 1862 cm2. Use the Pythagorean Theorem to find the length ofthe hypotenuse of the base of the triangular prism.

Triangular prism:

So, the surface area of four faces of the triangularprism is about 962.8 cm2. Therefore, the total surface area is about 1862 +

962.8 or 2824.8 cm2.

ANSWER:

2824.8 cm2


10-6 Surface Area

Find the surface area of each solid. Round to thenearest tenth.

38.

SOLUTION: We need to determine the slant height l. Usetrigonometry.

Use the exact value of l to find the lateral area.


ANSWER:

510.2 mm2

39.

SOLUTION: We need to determine the slant height and the

length of the base b. Use trigonometry. Slant height:

Base:

This value of x is the apothem, which is only half ofthe length of the sides of the base. Use the exact values of and x to find the lateralarea.

Now find the surface area.

ANSWER:

4524.9 ft2


10-6 Surface Area

40. CONSTRUCTION A road roller is a constructionvehicle with smooth and heavy rollers used forcompacting roads and pavement. One of these rollershas a diameter of 48 inches and is 36 inches in length.What is the area covered by the roller in two fullturns?

SOLUTION: Total area covered = 2(Lateral Area)

ANSWER:

10,857.3 in.2

41. SUNCATCHERS Abby makes suncatchers of glassto sell at art shows. One style of suncatcher is a righthexagonal prism with a height of 9 centimeters andeach base edge of 4 centimeters. What is the surfacearea of each suncatcher? (Hint: First, find the lengthof the apothem of the base.)

SOLUTION: Apothem of base of hexagon = 4sin 60º ≈ 3.464

ANSWER:

299.1 cm2

42. WRITING IN MATH A square-based prism and atriangular prism are the same height. The base of thetriangular prism is an equilateral triangle with analtitude equal in length to the side of the square.Compare the lateral areas of the prisms.

SOLUTION:

The lateral area of a prism is given by L = Ph. Let hrepresent the height of both prisms and s representthe length of a side of the square.The perimeter of the square is P = 4s, so the lateral

area of the square-based prism is given by L = (4s)h. The base is of the triangular prism is an equilateraltriangle with an altitude equal to the side of thesquare, or s. To find the perimeter of the triangle, firstfind the length of one of its sides.

Use the properties of the 30-60-90 right triangle tofind the length of the sides.

The side opposite the 60°-angle is times greaterthan the side opposite the 30°-angle. So, the sideopposite the 30°-angle is . The hypotenuse is

twice as long as the side opposite the 30°-angle or

. The perimeter of the equilateral triangle is

or , and the lateral area of the

triangular prism is . The two prisms have the same height. Compare theperimeters of their bases to compare their lateralareas. The perimeter of the square-based prism is greaterthan that of the triangular prism, since .Therefore, the lateral area of the square-based prismis greater than that of the triangular prism.

ANSWER:

The lateral area of the square-based prism is greaterthan that of the triangular prism. The square has aperimeter of 4s and the triangle has a perimeter of

and .

43. REASONING A cone and a square pyramid havethe same surface area. If the areas of their bases arealso equal, do they have the same slant height aswell? Explain.


10-6 Surface Area

SOLUTION:

The surface area of the cone is given by

where r is the radius of thebase and is the slant height. The surface area ofthe square pyramid is given by

where s is the length of each side of the square base

and is the slant height of the pyramid. Since thebase of the pyramid is a square, the perimeter is P =4s. The area of the base of the cone and the pyramid

are the same, so . The cone and the square pyramid have the samesurface areas, so set the two expressions equal.Subtract the area of the base from each side andthen solve for .

Use the equal base areas to find an equivalentexpression for s.

Substitute for s in the expression for .

The slant height of the cone is or about 1.13

times greater than the slant height of the pyramid.Therefore, they are not equal.

ANSWER:

They are not equal. The slant height of the cone is

or about 1.13 times greater than the slant

height of the square pyramid.

44. CRITIQUE ARGUMENTS Montell and Derekare finding the surface area of a cylinder with height5 centimeters and radius 6 centimeters. Is either ofthem correct? Explain.

SOLUTION:

Therefore, Derek is correct.

ANSWER:

Derek; sample answer: , so the

surface area of the cylinder is or

.

45. REASONING Classify the following statement assometimes, always, or never true. Justify yourreasoning.The surface area of a cone of radius r and heighth is less than the surface area of a cylinder ofradius r and height h.

SOLUTION:

Consider the cylinder below.

The surface area of this cylinder is 2πrh + 2πr2.


10-6 Surface Area

Now, consider a cone with the same base and height.

The surface area for the cone is πrl + πr2. Comparethe two formulas.

The values h, r, and form a triangle, so r + h mustbe greater than . Therefore, 2h + r is also greaterthan .Thus, the statement is always true.

ANSWER:

Always; if the heights and radii are the same, thesurface area of the cylinder will be greater since ithas two circular bases and additional lateral area.

46. ARGUMENTS Determine whether the followingstatement is true or false. Explain your reasoning.A square pyramid and a cone both have height hunits and base perimeter P units. Therefore, theyhave the same total surface area.

SOLUTION:

The surface area of a square pyramid is . The surface area of a cone is

. We know that the perimeter of the square base isequal to the perimeter(circumference) of the circularbase. We can use this information to get the side s interms of the radius r.

Now we can compare the areas of the bases.

The area of the circular base is greater than the areaof the square base. Now, we need to compare the lateral areas. Find . For the circle, the radius, the slant height, and theheight form a right triangle. For the square, half of theside, the slant height, and the height form a righttriangle. We have determined that the side of the square is , so half of the side is . This value is less than r, sowe know that the radius of the square is less than theradius of the circle, so the slant height of the cone isgreater than the slant height of the pyramid. Now, compare the lateral areas.


10-6 Surface Area

The slant height of the cone is greater than the slantheight of the pyramid, so the lateral area of the coneis greater than the lateral area of the pyramid. The lateral area and the base of the cone are greaterthan the lateral area and base of the pyramid, so thestatement is false.

ANSWER:

False; the lateral area and the base of the cone aregreater than the lateral area and base of the pyramid.

47. REASONING A right prism has a height of h unitsand a base that is an equilateral triangle of side units. Find the general formula for the total surfacearea of the prism. Explain your reasoning.

SOLUTION:

Draw the equilateral triangle. The altitude forms two30°-60°-90° triangles. The altitude is determined to

have a length of .

Find the area of the triangle.

The perimeter of the triangle is Find the surfacearea.

ANSWER:

the area of an equilateral triangle of side

is and the perimeter of the triangle is

So, the total surface area is

48. WRITING IN MATH Describe how to find thesurface area of a regular polygonal pyramid with an


10-6 Surface Area

n-gon base, height h units, and an apothem of a units.

SOLUTION:

Use the apothem, the height, and the PythagoreanTheorem to find the slant height of the pyramid.

Divide the regular n-gon for the base into congruentisosceles triangles. Each central angle of the n-gon

will have a measure of , so the measure of the

angle in the right triangle created by the apothem will

be ÷ 2 or . The apothem will bisect the

base of the isosceles triangle so if each side of theregular polygon is s, then the side of the right triangleis . Use a trigonometric ratio to find the length of aside s.

Then find the perimeter by using P = n × s.

Finally, use to find the surface area

where B is the area of the regular n-gon and is given

by

ANSWER:

Use the apothem, the height, and the PythagoreanTheorem to find the slant height of the pyramid.Then use the central angle of the n-gon and theapothem to find the length of one side of the n-gon.

Then find the perimeter. Finally, use to

find the surface area. The area of the base B is


10-6 Surface Area

49. Olivia makes a cylinder by bending the cardboardrectangle shown below so that the 8-centimeter sidesjoin to form the lateral face of the cylinder. Then shecuts out two cardboard circles to form the bases andattaches these to the lateral face.

Which of the following is the best estimate of thesurface area of the cylinder Olivia makes? A 26 cm²B 52 cm²C 144 cm²D 196 cm²

SOLUTION: Start by finding the radius of the circular bases of thecylinder that will be formed by rolling up thecardboard into a cylinder. Since the longer side of therectangle is the base edge, this dimension alsodoubles as the circumference of the circle. Use thisto approximate the radius of the circle:

Now, find the surface area of the newly formedcylinder, with a radius of 2.9 cm and a height of 8cm.

Therefore, the correct choice is D.

ANSWER: D

50. A cylindrical can has a circumference of 16π inchesand a height of 20 inches. What is the surface area ofthe can in square inches? Round to the nearest tenth.

SOLUTION: Start by finding the radius of the circular bases of thecylinder.

Now, find the surface area of the cylinder, with aradius of 8 in. and a height of 20 in.

ANSWER: 1407.4


10-6 Surface Area

51. DeMarco is wrapping presents for a party. Eachpresent is in a box, shaped like a rectangular prism,with the dimensions shown here. DeMarco plans towrap 8 of the boxes.

Which of the following is the best estimate for theleast amount of wrapping paper DeMarco will needto buy? A 1728 in²B 2016 in²C 2304 in²D 2592 in²

SOLUTION: Start by finding the surface area of one box:

Then, since Marcos wants to wrap 8 presents,multiply this surface area by 8:

The correct choice is C.

ANSWER: C

52. The top of a gazebo in a park is in the shape of aregular pentagonal pyramid. Each side of thepentagon is 10 feet long. If the slant height of the roofis about 6.9 feet, what is the lateral roof area to thenearest tenth?

SOLUTION: Sketch the pyramid, labeling it with the givendimensions:

The lateral area of the gazebo's roof is 172.5 ft².

ANSWER: 172.5


10-6 Surface Area

53. A model of a cone is used to demonstrate a new filterwith top. To the nearest square millimeter, what is thesurface area of the cone?

A 2705 mm2

B 3299 mm2

C 8820 mm2

D 9368 mm2

SOLUTION: Begin by using the Pythagorean Theorem to find theslant height of the cone.

Now, use the formula for a surface area of a cone,with a radius of 21 mm and a slant height of 29 mm.

The surface area of the cone is about 3299 mm². The correct choice is B.

ANSWER: B

54. MULTI-STEP There are two separate buildingsnext to each other. The first is in the shape of asquare prism. The dimensions of the base are 100feet by 100 feet and the height of the building is 22feet. The second building is in the shape of a squarepyramid with the same dimensions.

a. What is the surface area of the first building?

A 19,600 ft2

B 28,800 ft2

C 31,500 ft2

D 48,000 ft2

b. What is the surface area of the second building?

A 16,420 ft2

B 18,720 ft2

C 20,925 ft2

D 38,000 ft2

c. Why are the surface areas of the buildingsdifferent even though the dimensions are the same?

SOLUTION: a. The surface area of the first building is twice thearea of the base plus the lateral area.

The correct choice is B

b. The surface area of the second building is the areaof the base plus the lateral area.The slant height is not given, so use the PythagoreanTheorem

The correct choice is C. c. The prism has six faces and the pyramid has onlyfive. All of the faces of the prism are rectangles, butthe pyramid has four faces that are triangles and onlyone that is a rectangle.


10-6 Surface Area

ANSWER: a. B

b. C c. The prism has six faces and the pyramid has onlyfive. All of the faces of the prism are rectangles, butthe pyramid has four faces that are triangles and onlyone that is a rectangle.


10-6 Surface Area

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