# Surface Area of Pyramids and Cones Worksheet - Page 3

Surface Area of Pyramids and Cones Worksheet
• Page 3
21.
The base area of a cone is 452.16 cm.2 and its height is 16 cm. What is the surface area of the cone?
 a. 1205.76 cm2 b. 1215.76 cm2 c. 1226.41 cm 2 d. 1195.31 cm 2

#### Solution:

Base area of a cone = πr2 = 452.16 cm2 and its height = 16 cm.

3.14 × r2 = 452.16
[Substitute the value of π in πr2 = 452.16.]

3.14r23.14= 452.16 / 3.14
[Divide each side by 3.14.]

r2 = 144
[Simplify.]

r2 = 144
[Taking square root on both sides.]

r = 12
[Simplify.]

Slant height of the cone = r2+h2
[Formula.]

= 122+162
[Substitute the values.]

= 144+ 256
[Substitute the values of 122 and 162.]

= 400

= 20 cm
[Simplify.]

Slant height of the cone = 20 cm

Surface area of the cone = πrl + πr2
[Formula.]

= 3.14 × 12 × 20 + 3.14 × 122
[Substitute the values.]

= 753.60 + 452.16
[Multiply.]

= 1205.76

The surface area of the cone = 1205.76 cm2

22.
Find the surface area of a square pyramid. [$b$ = 5 yd; $h$ = 20 yd.]

 a. 225 yd2 b. 224 yd2 c. 220 yd2 d. 200 yd2

#### Solution:

The surface area of a square pyramid
= (area of the square + 4 × area of the lateral triangle)
[Since a square pyramid has 4 triangles and one square.]

= s × s + 4 × 1 / 2 × b × h
[Formula.]

= 5 × 5 + 4 × 1 / 2 × 5 × 20
[Substitute the values.]

= 25 + 200 = 225 yd2
[Simplify.]

Therefore, the surface area of the square pyramid is 225 yd2.

23.
Find the surface area of the cone to the nearest ten $\mathrm{ft}$2.
($d$ = 20 ft, $l$ = 39 ft)

 a. 1539 $\mathrm{ft}$2 b. 1,540 $\mathrm{ft}$2 c. 490 $\mathrm{ft}$2 d. 1539.09 $\mathrm{ft}$2

#### Solution:

Radius r of the base of the cone = d2 = 202 = 10 ft

The surface area of a cone = π r2 + π r l
[Formula.]

= π × 102 + π × 10 × 39
[Substitute the values.]

= 100 π + 390 π
[Simplify.]

= 490 π

= 1539.09 = 1540 ft2
[To the nearest ten.]

The surface area of a cone is 1539.09 and to the nearest ten is1540 ft2.

24.
The surface area of a square pyramid is given by 540 cm.2 and the side of the square is 10 cm. Find the slant height of the square pyramid.
 a. 20 cm b. 23 cm c. 22 cm d. 21 cm

#### Solution:

The surface area of the square pyramid = area of the square + 4 × area of the triangle
[Formula.]

Area of the square = s × s = 10 × 10 = 100 cm.2
[Substitute.]

Area of the triangle = 1 / 2× b × h = 1 / 2× 10 × h
[Substitute.]

540 = 100 + 4 × 1 / 2× 10 × h

540 = 100 + 4 × 5h
[Simplify.]

440 = 20h
[Subtract 100 from both the sides.]

h = 440 / 20= 22 cm.
[Divide by 20 on both the sides.]

So, the slant height of the square pyramid is 22 cm.

25.
The slant height of a strawberry cone is 5 in. and the diameter is 2.3 in. What is the total surface area of the strawberry cone? Let $\pi$ = 3.
 a. 22.21 in2. b. 22 in2. c. 21.21 in2. d. 21 in2.

#### Solution:

The slant height of the strawberry cone is 5 in. (i.e., l = 5 in.) and the diameter is 2.3 in.

radius = diameter / 2= 2.3 / 2= 1.15

The surface area of the strawberry cone = π r2 + π r l
[Formula.]

= 3 × 1.15 × 1.15 + 3 × 1.15 × 5
[Substitute.]

= 3.96 + 17.25 = 21.21 in.2

The surface area of the strawberry cone is 21.21 in.2.

26.
The curved surface area of a cone is 169.56 cm2 and its slant height is 54 cm. Find the total surface area of the cone. (Round the answer to the nearest hundredth unit.)
 a. 179.1 cm2 b. 172.7 cm2 c. 169.56 cm2 d. 223.56 cm2

#### Solution:

Curved surface area of cone = πrl
[Formula.]

Curved surface area of a cone = 169.56 cm2 and its slant height = 54 cm

πrl = 169.56
[From the steps-1 and 2.]

π × r × 54 = 169.56
[Substitute the value of l = 54.]

54 π r54 = 169.5654
[Divide each side by 54.]

π r = 3.14

π × r = 3.14

r = 3.14π

r = 1 cm
[Rounding the value.]

Total surface area of the cone = πrl + πr2
[Formula.]

= 169.56 + π × (1)2
[Substitute the value of π and curved surface area.]

= 172.7
[Simplify.]

The total surface area of the cone = 172.7 cm2

27.
What is the surface area of a pyramid?
 a. Twice the lateral area of the pyramid + base area of the pyramid b. Lateral area of the pyramid + base area of the pyramid c. Lateral area of the pyramid + twice the base area of the pyramid d. None of the above

#### Solution:

Surface area is sum of the areas of all faces.

Pyramids have only one base.

Surface area of a pyramid = lateral area of the pyramid + base area of the pyramid.

28.
The total surface area of a cone is 60 in.2 If its slant height is four times the radius, then what is the base diameter of the cone? [Use $\pi$ = 3.]

 a. 7 in. b. 8 in. c. 4 in. d. 10

#### Solution:

The total surface area of a cone = πrl + πr2 = 60 in.2

Slant height l = 4 × radius = 4r

3 × r × 4 r + 3 × r2 = 60
[Substitute l = 4r and π = 3.]

12r2 + 3r2 = 60
[Multiply.]

15r2 = 60

15r215 = 60 / 15
[Divide each side by 15.]

r2 = 4
[Simplify.]

r2 = 4
[Take square root on both sides.]

r = 2
[Simplify.]

Base radius of the cone = 2 in.

Base diameter of the cone = 2 × radius = 2 × 2 = 4 in.

29.
A circular cone with a base radius of 1 ft has a surface area of 15.7 ft2. What is its slant height?
 a. 14 ft b. 9 ft c. 24 ft d. 4 ft

#### Solution:

Surface area of a cone = π rl + πr2 = 15.7 ft2

3.14 x 1 x l + 3.14 x 12 = 15.7
[Substitute the values of π and of the radius.]

3.14 l + 3.14 = 15.7
[Multiply.]

3.14 l + 3.14 - 3.14 = 15.7 - 3.14
[Subtract 3.14 from each side.]

3.14 l = 12.56
[Simplify.]

l = 4
[Divide each side by 3.14.]

The slant height of the cone is 4 ft .

30.
Find the surface area of the rectangular pyramid.

 a. 428 m2 b. 564 m2 c. 550 m2 d. 496 m2

#### Solution:

From the figure, the length and the width of the base rectangle are 18 m and 10 m

Area of the lateral face along the length of the rectangle = 1 / 2 × base × height
= 1 / 2 × 18 × 13 = 117 m2
[Lateral face is a triangle with base 18 m and height 13 m.]

Area of the lateral face along the width of the rectangle = 1 / 2 × base × height
= 1 / 2 × 10 × 15 = 75 m2
[Lateral face is a triangle with base 10 m and height 15 m.]

Surface area of the rectangular pyramid = base area + lateral area
[Formula.]

= base area + 2 × area of lateral face along length of the base rectangle + 2 × area of lateral face along the width of the base rectangle

= (18 × 10) + 2(117) + 2(75)
[Substitute the values.]

= 180 + 234 + 150
[Multiply.]

= 564