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# 10 6 Surface Area

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```10-6 Surface Area
Find the surface area of each solid. Round to the
nearest tenth if necessary.
1.
3.
SOLUTION:
The surface area S of a regular
pyramid is
, where
SOLUTION:
The radius is 7.5 cm and the height is 15 cm.
L is the lateral area and B is the area of the base.
Use the Pythagorean Theorem to find the slant
height.
Therefore, the surface area of the pyramid is 640
cm2.
ANSWER:
640 cm2
Find the surface area.
2.
SOLUTION:
Surface area = 2(10 m)(11 m) + 2(10 m)(15 m) +
2(11 m)(15 m) = 850 m&sup2;
ANSWER:
≈ 571.9 cm2
ANSWER:
850 m&sup2;
4.
SOLUTION:
The segment joining the points where the slant height
and height intersect the base is the apothem.
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10-6 Surface Area
5.
Use the Pythagorean Theorem to find the length the
apothem of the base.
SOLUTION:
The base of the prism is a right triangle with the legs
8 ft and 6 ft long. Use the Pythagorean Theorem to
find the length of the hypotenuse of the base.
A central angle of the regular hexagon is
so the angle formed in the triangle below is 30.
,
Find the surface area.
Use a trigonometric ratio to find the length of a side
of the hexagon s.
ANSWER:
336 ft2
Find the surface area of the pyramid.
Therefore, the surface area of the pyramid is about
332.6 m2.
ANSWER:
≈ 332.6 m2
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10-6 Surface Area
8. PATIO STONES A patio stone has a rectangular
base that is 3 inches by 8 inches and a height of 4
inches. What is the surface area of the stone?
6.
SOLUTION:
SOLUTION:
ANSWER:
≈ 2063.6 cm2
7. CARS Evan is buying new tire rims that are 14
inches in diameter and 6 inches wide. Determine the
surface area of each rim. Round to the nearest tenth.
SOLUTION:
ANSWER:
571.8 in2
ANSWER:
9. ROOFING A pyramid shaped roof has a square
base that is 30 feet wide and a slant height of 14 feet.
How much roofing material is needed to cover the
roof?
SOLUTION:
ROOFING A pyramid shaped roof has a square
base that is 30 feet wide and a slant height of 14 feet.
How much roofing material is needed to cover the
roof?
The amount of roofing material needed to cover the
roof is the lateral area of the pyramid.
L = (0.5)Pl
= (0.5)(4)(30 ft)(14 ft)
= 840 ft&sup2;
ANSWER:
840 ft&sup2;
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10-6 Surface Area
10. TANKS A storage tank is shown at the right.
Find the surface area of each solid. Round to the
nearest tenth if necessary.
Round answers to the nearest tenth.
11.
SOLUTION:
Find the length of the third side of the triangle.
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.
Now find the surface area.
b. Find the lateral area of a cone with a radius of 7
feet and a slant height of 9 feet.
ANSWER:
36 ft2
c. Find the total lateral area of the tank by adding
the lateral area of the cylinder and the lateral area of
the cone.
ANSWER:
a. 483.8 ft&sup2;
b. 197.9 ft&sup2;
c. 681.7 ft&sup2;
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10-6 Surface Area
13.
12.
SOLUTION:
The diameter of the base is 3.6 cm and the height of
the cylinder is 1.1 cm.
SOLUTION:
The radius of the cone is
and the height
is 18 cm. Use the Pythagorean Theorem to find the
slant height .
The total surface area of the prism is the sum of the
areas of the bases and the lateral surface area.
ANSWER:
≈ 32.8 cm2
Find the surface areas of the cone.
Therefore, the surface area of the cone is about
470.7 cm2.
14.
SOLUTION:
ANSWER:
≈ 470.7 cm2
We need to find the area of the triangle to determine
the area of the bases. Use the Pythagorean Theorem
to find the height of the triangles.
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10-6 Surface Area
15.
SOLUTION:
The base of the pyramid is a square with side length
of 8 feet and the height is 10 feet. Use the
Pythagorean Theorem to find the slant height of the
pyramid.
Use
to calculate the surface area.
Find the surface area of the pyramid.
ANSWER:
151.9 m2
Therefore, the surface area of the pyramid is about
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
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10-6 Surface Area
Theorem to find the length of the apothem of the
base.
Find the surface area of the pyramid.
Therefore, the surface area of the pyramid is about
302.9 cm2.
ANSWER:
≈ 302.9 cm2
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&deg;.
17. A cone has a diameter of 3.4 centimeters, and a slant
height of 6.5 centimeters.
SOLUTION:
The radius of the cone is
.
Find the surface area.
Use a trigonometric ratio to find the length s of each
side of the triangle.
Therefore, the surface area of the cone is about 43.8
cm2.
ANSWER:
≈ 43.8 cm2
Use the formulas for regular polygons to find the
perimeter and area of the base.
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10-6 Surface Area
18. A rectangular prism has = 25 centimeters, w = 18
centimeters, and h = 12 centimeters.
SOLUTION:
Note that the surface area of the solid is the same no
matter which face is the base.
So, the area of the base B is about 93.5 mm2.
ANSWER:
Find the surface area of the pyramid.
1932 cm2
19. A regular hexagonal pyramid has a base edge of 6
millimeters and a slant height of 9 millimeters.
SOLUTION:
Therefore, the surface area of the pyramid is about
The base of the pyramid is a regular hexagon. The
perimeter of the hexagon is P = 6 &times; 6 or 36 mm.
255.5 mm2.
ANSWER:
≈ 255.5 mm2
A central angle of the hexagon is
angle formed in the triangle below is 30&deg;.
, so the
Use a trigonometric ratio to find the apothem a and
then find the area of the base.
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10-6 Surface Area
20. A triangular prism has h = 6 inches and a right
triangle base with legs 9 inches and 12 inches.
SOLUTION:
Find the other side of the triangular base.
Now you can find the urface area.
A central angle of the square is
, so the
angle formed in the triangle below is 45&deg;.
ANSWER:
324 in2
21. A cylinder has a diameter of 8 inches and a height of
6.2 inches.
Use a trigonometric ratio to find the length of each
side of the square.
SOLUTION:
ANSWER:
≈ 256.4 in.2
Find the perimeter and area of the base.
22. A square pyramid has an altitude of 12 inches and a
slant height of 18 inches.
SOLUTION:
The base of the pyramid is square. Use the
Pythagorean Theorem to find the length the apothem
of the base.
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Find the surface area of the pyramid.
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10-6 Surface Area
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
Therefore, the surface area of the pyramid is about
1686.0 in2.
is
or 9.5 feet. Use the Pythagorean Theorem to
find the radius.
ANSWER:
≈ 1686.0 in2
23. A cylinder has a radius of 3 millimeters and a height
of 15 millimeters.
SOLUTION:
ANSWER:
Find the surface area of the cone.
≈ 339.3 mm2
Therefore, the surface area of the cone is about
446.1 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.
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10-6 Surface Area
The cone has a radius of 5 feet and a height of 6
feet. Use the Pythagorean Theorem to find the slant
height of the cone.
26. Find the lateral area of the dog house with a 12 inch
square 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 of
the door:
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 and
cylinder to find the lateral area of the tent.
The lateral area of the walls minus the door is
.
Next, find the lateral area of the roof (square
pyramid):
Therefore, the lateral area of the tent is about 311.2
ft2.
ANSWER:
311.2 ft2
The total lateral area of the dog house is the
combination of the walls ( minus the door) and the
roof. Combine the lateral areas of each to get the
total lateral area:
ANSWER:
6465.1 in&sup2;
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10-6 Surface Area
Find the surface area of each composite solid.
Round to the nearest tenth if necessary.
28.
27.
SOLUTION:
The solid is a combination of a rectangular prism and
a cylinder. The base of the rectangular prism is 6 in
by 4 in and the radius of the cylinder is 3 in. The
height of the solid is 15 in.
SOLUTION:
The solid is a combination of a cube and a cylinder.
The length of each side of the cube is 12 cm and the
radius of the cylinder is 6 cm. The height of the solid
is 12 cm.
Rectangular prism:
Rectangular prism:
The surface area of five faces of the rectangular
prism is 720 cm2.
The surface area of five faces of the rectangular
prism is 258 cm2.
Half-cylinder:
Half-cylinder:
The surface area of the half-cylinder is 108π cm2 and
the total surface area is about 1059.3 cm2.
The surface area of the half-cylinder is 54π
cm2.
ANSWER:
The total surface area is 258 + 54π = 427.6.
1059.3 cm2
ANSWER:
427.6 in2
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10-6 Surface Area
29. MOUNTAINS A conical mountain has a radius of
1.6 kilometers and a height of 0.5 kilometer. What is
the lateral area of the mountain?
SOLUTION:
The radius of the conical mountain is 1.6 kilometers
and the height is 0.5 kilometers. Use the Pythagorean
Theorem to find the slant height.
30. AQUARIUMS The Tower Aquarium in Henley
Beach, Australia, is the world’s largest cylindrical
aquarium. It reaches a height of over 40 meters and
is 36 meters in diameter. Visitors ascend through a
column of water as they ride a split-level glass lift up
seven floors, through the center of the aquarium.
What is the approximate lateral area of the outside
of the aquarium?
SOLUTION:
Since the diameter of the cylinder is 36 m, the radius
is
.
Find the lateral area L of the conical mountain.
ANSWER:
Therefore, the lateral area is about 8.4 km2.
ANSWER:
8.4 km2
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10-6 Surface Area
31. HISTORY Archaeologists recently discovered a
1500-year-old pyramid in Mexico City. The square
pyramid measures 165 yards on each side and once
stood 20 yards tall. What was the original lateral area
of the pyramid?
SOLUTION:
The pyramid has a square base with sides having
lengths of 165 yards and a height of 20 yards. Use
32. MONUMENTS The monolith mysteriously
appeared overnight at Seattle, Washington’s
Manguson Park. It is a hollow rectangular prism 9
feet tall, 4 feet wide, and 1 foot thick.
a. Find the area in square feet of the structure’s
surfaces that lie above the ground.
b. Use dimensional analysis to find the area in square
yards.
the Pythagorean Theorem to find the slant height.
SOLUTION:
a. The area of surfaces that lie above the ground is
the sum of the area of the upper base and the lateral
surface area.
b.
Find the lateral area L of a regular pyramid.
Therefore, the lateral area of the pyramid is about
28,013.6 yd2.
ANSWER:
a. 94 ft2
b. 10.4 yd2
ANSWER:
28,013.6 yd2
33. TEPEES The dimensions of two canvas tepees are
shown in the table at the right. Including the floors,
approximately how much more canvas is used to
make Tepee B than Tepee A?
SOLUTION:
The tepees are in the shape of a right cone. To find
the amount of canvas used for each tepee, we need
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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 B will use about 380.1 ft2 more canvas than
Tepee A.
ANSWER:
about 380.1 ft2
34. DESIGN A mailer needs to hold a poster that is
almost 38 inches long and has a maximum rolled
diameter of 6 inches.
a. Design a mailer that is a triangular prism. Sketch
the mailer and its net.
b. Suppose you want to minimize the surface area of
the mailer. What would be the dimensions of the
mailer and its surface area?
Tepee A
Tepee B
SOLUTION:
a. A triangular prism should consist of two triangles
and three rectangles. They should be connected so
that, when folded together they form a prism.
Use the Pythagorean Theorem to find the slant height
of each tepee and then find the surface area for
each cone.
b. In order to minimize the surface area of the
triangular prism, the triangles should be equilateral,
and the side lengths of the rectangles should coincide
the the length of the base of the triangle and the
length of the poster. The surface area will then be
.
To find how much more canvas is used to make
Tepee B than Tepee A, subtract the surface areas.
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Use trigonometry to find the area of the triangles.
The diameter of the poster has a maximum of 6 in.
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10-6 Surface Area
which corresponds to a radius of 3 in.
The total surface area can be calculated:
ANSWER:
a. Sample answer:
For the base we have:
For the height we have:
b. side lengths of triangular bases, about 10.39
in.each; height, 38 in.; 1278 in2
Now calculate the area:
The area of the rectangles will be the product of the
area of the base of the triangle with the length of the
poster.
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35. MULTI-STEP Hector is designing a glass
greenhouse for a city park. He has a 40-foot by 20foot rectangular plot available. He wants the roof to
be a triangular prism in which the center of the roof is
4 feet higher than the edges. The glass costs \$25 per
square foot, and Hector cannot spend more than
\$60,000 on glass.
a. What is the maximum height that Hector should
make 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 given
information, with g representing the height of the
roof.
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10-6 Surface Area
after the top portion has been cut by a plane parallel
to the base. The ferret tent shown is a frustum of a
regular pyramid.
a. Describe the faces of the solid.
b. Find the surface area of the frustum formed by the
tent.
c. Another pet tent is made by cutting the top half off
of a pyramid with a height of 12 centimeters, slant
height of 20 centimeters and square base with side
lengths of 32 centimeters. Find the surface area of
the frustum.
Find the sum of the surface areas of each individual
section. The rectangular section of the front and back
is 2 &times; 20 &times; g or 40g ft&sup2;. The sides cover 2 &times; 40 &times; g
or 80g ft&sup2;. The triangular tops of the front and back
of the greenhouse cover 2(0.5)(4)(20) or 80 ft&sup2;. The
slant of the roof is
.
Thus, the roof covers 2(40)(10.77) or 861 ft&sup2;.
The total surface area is 861 + 80 + 120g ft&sup2;.
Hector can use up to 60,000 &divide; 25 or 2400 ft&sup2;.
Therefore, g is approximately 12.1. Rounding down,
we get a height of 12 ft.
c. Hector used the entire available plot. There was no
glass used for the base. The entrance was made of
glass. The top of the roof ran along the 40 ft length of
the greenhouse.
ANSWER:
a. about 12 ft
b. First, find the sum of the surface areas of each
individual section. The rectangular section of the front
and back is 2 &times; 20 &times; g or 40g ft&sup2;. The sides cover 2 &times;
40 &times; g or 80g ft&sup2;. The triangular tops of the front and
back of the greenhouse cover 2(0.5)(4)(20) or 80 ft&sup2;.
The slant of the roof is
. Thus, the roof
covers 2(40)(10.77) or 861 ft&sup2;. The total surface area
is 861 + 80 + 120g ft&sup2;. Hector can use up to 60,000 &divide;
25 or 2400 ft&sup2;. Therefore, g is approximately 12.1.
Rounding down, we get a height of 12 ft.
c. Hector used the entire available plot. There was no
glass used for the base. The entrance was made of
glass. The top of the roof ran along the 40 ft length of
the greenhouse.
SOLUTION:
a. The two bases are squares and the 4 lateral faces
are 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 trapezoid
with bases b 1, b 2 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 12
cm, slant height of 20 cm and square base with side
lengths of 32 cm is cut, the height of the frustum will
be 6 cm, the slant height 10 cm and the length of
each side of the upper base will be 16 cm.
36. PETS A frustum is the part of a solid that remains
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10-6 Surface Area
So, the surface area of five faces of the rectangular
prism is 1862 cm2.
The total surface area of the frustum will be
Use the Pythagorean Theorem to find the length of
the hypotenuse of the base of the triangular prism.
ANSWER:
a. 4 trapezoids, 2 squares
b. 1015 in2
Triangular prism:
c. 2240 cm2
37. The three-dimensional box needs to have a clear
coating painted on all six faces. What is the
approximate surface area of the box?
So, the surface area of four faces of the triangular
prism is about 962.8 cm2.
Therefore, the total surface area is about 1862 +
962.8 or 2824.8 cm2.
SOLUTION:
This composite solid can be divided into a rectangular
prism 13 cm by 21 cm by 28 cm and a triangular
prism that has a right triangle with legs of 7 cm and
21 cm as the base and a height of 28 cm. The
surface area of the solid is the sum of the surface
areas of each prism without the area of the 21 cm by
28 cm rectangular face at which they are joined.
ANSWER:
2824.8 cm2
.
Rectangular prism:
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10-6 Surface Area
Find the surface area of each solid. Round to the
nearest tenth.
length of the base b. Use trigonometry.
Slant height:
38.
SOLUTION:
We need to determine the slant height l. Use
trigonometry.
Use the exact value of l to find the lateral area.
Base:
This value of x is the apothem, which is only half of
the length of the sides of the base.
Use the exact values of
area.
and x to find the lateral
Find the surface area.
Now find the surface area.
ANSWER:
510.2 mm2
ANSWER:
4524.9 ft2
39.
SOLUTION:
We need to determine the slant height
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and the
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10-6 Surface Area
40. CONSTRUCTION A road roller is a construction
vehicle with smooth and heavy rollers used for
compacting roads and pavement. One of these rollers
has a diameter of 48 inches and is 36 inches in length.
What is the area covered by the roller in two full
turns?
area of the square-based prism is given by L = (4s)h.
The base is of the triangular prism is an equilateral
triangle with an altitude equal to the side of the
square, or s. To find the perimeter of the triangle, first
find the length of one of its sides.
SOLUTION:
Total area covered = 2(Lateral Area)
Use the properties of the 30-60-90 right triangle to
find the length of the sides.
ANSWER:
10,857.3 in.2
41. SUNCATCHERS Abby makes suncatchers of glass
to sell at art shows. One style of suncatcher is a right
hexagonal prism with a height of 9 centimeters and
each base edge of 4 centimeters. What is the surface
area of each suncatcher? (Hint: First, find the length
of the apothem of the base.)
SOLUTION:
Apothem of base of hexagon = 4sin 60&ordm; ≈ 3.464
The side opposite the 60&deg;-angle is
times greater
than the side opposite the 30&deg;-angle. So, the side
opposite the 30&deg;-angle is
. The hypotenuse is
twice as long as the side opposite the 30&deg;-angle or
. The perimeter of the equilateral triangle is
or
triangular prism is
, and the lateral area of the
.
The two prisms have the same height. Compare the
perimeters of their bases to compare their lateral
areas.
ANSWER:
299.1 cm2
42. WRITING IN MATH A square-based prism and a
triangular prism are the same height. The base of the
triangular prism is an equilateral triangle with an
altitude 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 h
represent the height of both prisms and s represent
the length of a side of the square.
The perimeter of the square is P = 4s, so the lateral
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The perimeter of the square-based prism is greater
than that of the triangular prism, since
.
Therefore, the lateral area of the square-based prism
is greater than that of the triangular prism.
ANSWER:
The lateral area of the square-based prism is greater
than that of the triangular prism. The square has a
perimeter of 4s and the triangle has a perimeter of
and
.
43. REASONING A cone and a square pyramid have
the same surface area. If the areas of their bases are
also equal, do they have the same slant height as
well? Explain.
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10-6 Surface Area
SOLUTION:
The surface area of the cone is given by
where r is the radius of the
base and
is the slant height. The surface area of
the 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 the
base 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
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 Derek
are finding the surface area of a cylinder with height
5 centimeters and radius 6 centimeters. Is either of
them correct? Explain.
.
The cone and the square pyramid have the same
surface areas, so set the two expressions equal.
Subtract the area of the base from each side and
then solve for .
SOLUTION:
Therefore, Derek is correct.
ANSWER:
Use the equal base areas to find an equivalent
expression for s.
Substitute for s in the expression for
.
Derek; sample answer:
surface area of the cylinder is
, so the
or
.
45. REASONING Classify the following statement as
sometimes, always, or never true. Justify your
reasoning.
The surface area of a cone of radius r and height
h is less than the surface area of a cylinder of
radius r and height h.
SOLUTION:
Consider the cylinder below.
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.
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The surface area of this cylinder is 2πrh + 2πr2.
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10-6 Surface Area
Now, consider a cone with the same base and height.
Now we can compare the areas of the bases.
The surface area for the cone is πrl + πr2. Compare
the two formulas.
The values h, r, and form a triangle, so r + h must
be greater than . Therefore, 2h + r is also greater
than .
Thus, the statement is always true.
ANSWER:
Always; if the heights and radii are the same, the
surface area of the cylinder will be greater since it
has two circular bases and additional lateral area.
46. ARGUMENTS Determine whether the following
statement is true or false. Explain your reasoning.
A square pyramid and a cone both have height h
units and base perimeter P units. Therefore, they
have the same total surface area.
SOLUTION:
The surface area of a square pyramid is
. The surface area of a cone is
The area of the circular base is greater than the area
of the square base.
Now, we need to compare the lateral areas. Find .
For the circle, the radius, the slant height, and the
height form a right triangle. For the square, half of the
side, the slant height, and the height form a right
triangle.
We have determined that the side of the square is
, so half of the side is
. This value is less than r, so
we know that the radius of the square is less than the
radius of the circle, so the slant height of the cone is
greater than the slant height of the pyramid.
Now, compare the lateral areas.
.
We know that the perimeter of the square base is
equal to the perimeter(circumference) of the circular
base. We can use this information to get the side s in
terms of the radius r.
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10-6 Surface Area
47. REASONING A right prism has a height of h units
and a base that is an equilateral triangle of side
units. Find the general formula for the total surface
area of the prism. Explain your reasoning.
SOLUTION:
Draw the equilateral triangle. The altitude forms two
30&deg;-60&deg;-90&deg; triangles. The altitude is determined to
have a length of
.
The slant height of the cone is greater than the slant
height of the pyramid, so the lateral area of the cone
is greater than the lateral area of the pyramid.
The lateral area and the base of the cone are greater
than the lateral area and base of the pyramid, so the
statement is false.
ANSWER:
False; the lateral area and the base of the cone are
greater than the lateral area and base of the pyramid.
Find the area of the triangle.
The perimeter of the triangle is
area.
Find the surface
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 the
surface area of a regular polygonal pyramid with an
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10-6 Surface Area
n-gon base, height h units, and an apothem of a units.
SOLUTION:
Use the apothem, the height, and the Pythagorean
Theorem to find the slant height of the pyramid.
Then find the perimeter. Finally, use
to
find the surface area. The area of the base B is
Divide the regular n-gon for the base into congruent
isosceles 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
&divide; 2 or
. The apothem will bisect the
base of the isosceles triangle so if each side of the
regular polygon is s, then the side of the right triangle
is . Use a trigonometric ratio to find the length of a
side s.
Then find the perimeter by using P = n &times; 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 Pythagorean
Theorem to find the slant height of the pyramid.
Then use the central angle of the n-gon and the
apothem to find the length of one side of the n-gon.
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10-6 Surface Area
49. Olivia makes a cylinder by bending the cardboard
rectangle shown below so that the 8-centimeter sides
join to form the lateral face of the cylinder. Then she
cuts out two cardboard circles to form the bases and
attaches these to the lateral face.
Which of the following is the best estimate of the
surface area of the cylinder Olivia makes?
A 26 cm&sup2;
B 52 cm&sup2;
C 144 cm&sup2;
D 196 cm&sup2;
SOLUTION:
Start by finding the radius of the circular bases of the
cylinder that will be formed by rolling up the
cardboard into a cylinder. Since the longer side of the
rectangle is the base edge, this dimension also
doubles as the circumference of the circle. Use this
to approximate the radius of the circle:
50. A cylindrical can has a circumference of 16π inches
and a height of 20 inches. What is the surface area of
the can in square inches? Round to the nearest tenth.
SOLUTION:
Start by finding the radius of the circular bases of the
cylinder.
Now, find the surface area of the cylinder, with a
radius of 8 in. and a height of 20 in.
ANSWER:
1407.4
Now, find the surface area of the newly formed
cylinder, with a radius of 2.9 cm and a height of 8
cm.
Therefore, the correct choice is D.
ANSWER:
D
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10-6 Surface Area
51. DeMarco is wrapping presents for a party. Each
present is in a box, shaped like a rectangular prism,
with the dimensions shown here. DeMarco plans to
wrap 8 of the boxes.
52. The top of a gazebo in a park is in the shape of a
regular pentagonal pyramid. Each side of the
pentagon is 10 feet long. If the slant height of the roof
is about 6.9 feet, what is the lateral roof area to the
nearest tenth?
SOLUTION:
Sketch the pyramid, labeling it with the given
dimensions:
Which of the following is the best estimate for the
least amount of wrapping paper DeMarco will need
to buy?
A 1728 in&sup2;
B 2016 in&sup2;
C 2304 in&sup2;
D 2592 in&sup2;
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.
The lateral area of the gazebo's roof is 172.5 ft&sup2;.
ANSWER:
172.5
ANSWER:
C
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10-6 Surface Area
53. A model of a cone is used to demonstrate a new filter
with top. To the nearest square millimeter, what is the
surface area of the cone?
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 2705 mm2
A 16,420 ft2
B 3299 mm2
B 18,720 ft2
C 8820 mm2
C 20,925 ft2
D 9368 mm2
D 38,000 ft2
SOLUTION:
Begin by using the Pythagorean Theorem to find the
slant height
of the cone.
c. Why are the surface areas of the buildings
different even though the dimensions are the same?
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 correct choice is B
SOLUTION:
a. The surface area of the first building is twice the
area of the base plus the lateral area.
b. The surface area of the second building is the area
of the base plus the lateral area.
The slant height is not given, so use the Pythagorean
Theorem
The surface area of the cone is about 3299 mm&sup2;.
The correct choice is B.
ANSWER:
B
54. MULTI-STEP There are two separate buildings
next to each other. The first is in the shape of a
square prism. The dimensions of the base are 100
feet by 100 feet and the height of the building is 22
feet. The second building is in the shape of a square
pyramid with the same dimensions.
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The correct choice is C.
c. The prism has six faces and the pyramid has only
five. All of the faces of the prism are rectangles, but
the pyramid has four faces that are triangles and only
one that is a rectangle.
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10-6 Surface Area
ANSWER:
a. B
b. C
c. The prism has six faces and the pyramid has only
five. All of the faces of the prism are rectangles, but
the pyramid has four faces that are triangles and only
one that is a rectangle.
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