﻿ Surface Area of Pyramids and Cones Worksheet - Page 2 | Problems & Solutions
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Surface Area of Pyramids and Cones Worksheet
• Page 2
11.
A circular cone is 12 in. high. The radius of the base is 16 in. What is the lateral surface area of the cone?[Lateral area of cone = π × $r$ × $l$, where $r$, $l$ are the radius and height of the cone]  a. 28 in.2 b. 192 in.2 c. 320π in.2 d. 328.6 in.2

#### Solution:

Height of a circular cone, h = 12 in. and its base radius, r = 16 in.

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

= 122+162
[Substitute the values.]

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

= 400
[Add.]

= 20
[Simplify.]

Slant height of the cone = 20 in.

lateral surface area of the cone = πrl
[Formula.]

= π × 16 × 20
[Substitute the values.]

= 320π in.2

The lateral surface area of the cone = 320π in.2

Correct answer : (3)
12.
The diameter of an ice cream cone is 4 cm and the slant height is 9 cm. What is the surface area of the cone? Use π = 3.14. a. 75.48 cm2 b. 69.08 cm2 c. 65.08 cm2 d. 73.08 cm2

#### Solution:

Diameter of an ice cream cone = 4 cm and its slant height, l = 9 cm

Radius of an ice cream cone, r = diameter / 2 = 4 / 2 = 2 cm
[Substitute diameter = 4.]

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

= π × 2 × 9 + π × 22
[Substitute the values.]

= 18π + 4π

= 22π
[Add.]

= 22 × 3.14
[Substitute the value of π = 3.14.]

= 69.08
[Multiply.]

The surface area of the cone = 69.08 cm2

Correct answer : (2)
13.
What length should a 5 m wide canvas be cut in order to make a conical tent of 12 m base diameter and 4.3 m slant height? Find the cost of the tent if the canvas costs $12 per meter. (Round the answer to the nearest hundredth of a unit.) a. length = 16.21 m, cost =$194.52 b. length = 16.21 m, cost = $198.52 c. length = 20.21 m, cost =$198.52 d. length = 12.21 m, cost = $190.52 #### Solution: Diameter of the conical tent = 12 m and its slant height, l = 4.3 m Radius of the tent, r = diameter / 2 = 12 / 2 = 6 m Lateral area of the tent = πrl [Formula.] = π × 6 × 4.3 [Substitute the values.] = 81.09 m2 [Multiply.] Lateral area of the tent = 81.09 m2 Required canvas = lateral area of the tent = 81.09 m2 Width of the canvas = 5 m Length of the canvas = area of canvaswidth of canvas = 81.095 [Substitute the values.] = 16.21 m Length of the canvas = 16.21 m Cost of the canvas =$12 per meter.

Total cost of required canvas = 16.21 × 12 = $194.52 The length of the canvas is 16.21 m and it's total cost is$194.52

Correct answer : (1)
14.
A paper cone is 60 in. high and has a radius of 25 in.. Find the area of the paper needed to make the cone. a. 7135 in.2 b. 7087 in.2 c. 7065 in.2 d. 7100 in.2

#### Solution:

Height of the cone, h = 60 in. and its base radius, r = 25 in.

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

= 602+252
[Substitute the values.]

= 3600+625
[Substitute the values of 602 and 252.]

= 4225
[Add.]

= 65
[Simplify.]

Slant height of the cone, l = 65 in.

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

= π × 25 × 65 + π × 252
[Substitute the values.]

= 90π + 625π
[Multiply.]

= 2250π
[Add.]

= 2250 × π

= 7065 in.2
[Multiply.]

The surface area of the cone = 7065 in.2

Area of the paper needed = surface area of the cone = 7065 in.2

The area of the paper needed to construct the cone is 7065 in.2

Correct answer : (3)
15.
You need to paint the out side of the house as shown in the figure. A gallon of paint covers about 1296 square feet. How many gallons will you need to buy? (Ignore the areas of window and door.)[Given $h$ = 18, $b$ = 36.]  a. 6 gallons b. 4 gallons c. 3 gallons d. 5 gallons

#### Solution:

From the figure, bottom represent cube and top is a pyramid. The total shape of a house is a cubical pyramid.

Out side area of the house = 4 × area of square + 4 × area of triangle
[Formula.]

Length of the side of a square in the cube = 36 ft

Area of the square = side × side = 36 × 36 = 1296 ft2

Base of the triangle = 36 ft and its height = 18 ft

Area of the triangle = 1 / 2 × base × height = 1 / 2 × 36 × 18 = 324 ft2

Out side area of the house = 4 × 1296 + 4 × 324
[Substitute the values in the step-2.]

= 6480
[Add.]

Out side area of the house = 6480 ft2

A gallon of paint covers the area = 1296 ft2

Number of gallons required to cover the area of the out side of the house = 6480 / 1296= 5.

The number of gallons required to paint the out side of the house = 5.

Correct answer : (4)
16.
Find the length of paper of width 30 in. required to make a hollow cone of radius 18 in. and slant height 45 in. a. 79.85 in. b. 89.85 in. c. 94.85 in. d. 84.85 in.

#### Solution:

Radius of a cone = 18 in. and its slant height = 45 in.

Lateral area of the cone = πrl
[Formula.]

= π × 18 × 45
[Substitute the values.]

= 2545.71 in.2
[Multiply.]

The lateral area of the cone = 2545.71 in.2

Area of the required paper = lateral area of the cone = 2545.71 in.2

Width of the paper = 30 in.

Length of the paper = area of the paperwidth of the paper
[Formula.]

= 2545.7130
[Substitute the values.]

= 84.85 in.
[Simplify.]

The length of the paper = 84.85 in.

Correct answer : (4)
17.
The slant height($l$) of a cone is 40 in.. and its base radius($r$) is 32 in.. What is its height? a. 44 in. b. 24 in. c. 34 in. d. 54 in.

#### Solution:

Slant height of a cone = 40 in.. and its base radius = 32 in.

Let h = height of the cone.

Slant height = h2+r2
[Formula.]

40 = h2+322
[Substitute the values.]

402 = (h2+322)2
[Squaring on both sides.]

1600 = h2 + 1024
[Simplify.]

1600 - {1024 }= h2 + 1024 - 1024
[Subtract 1024 on both sides.]

576 = h2
[Simplify.]

576 = h2
[Taking square root on both sides.]

24 = h
[Simplify.]

The height of the cone = 24 in.

Correct answer : (2)
18.
The curved surface area of a cone is 75$\pi$ sq. units, whose slant height is 3 times the radius of its base. Find the slant height of the cone. a. 13 units b. 17 units c. 19 units d. 15 units

#### Solution:

Curved surface area of a cone = πrl = 75π.

Let r = base radius of the cone.

Slant height of the cone (l) = 3 × radius = 3r.

π × r × 3r = 75π
[Substitute the value of l = 3r in πrl = 75π.]

3πr2 = 75π
[Multiply.]

3πr23π = 75π3π
[Divide each side by 3π.]

r2 = 25
[Simplify.]

r2 = 25
[Taking square root on both sides.]

r = 5
[Simplify.]

Slant height of the cone = 3r = 3 × 5 = 15 units

The slant height of the cone = 15 units.

Correct answer : (4)
19.
What is the base area of a cone with radius 7 in. and height 29 in.? a. 153.86 in.2 b. 164.73 in.2 c. 174.73 in.2 d. 142.99 in.2

#### Solution:

Radius of a cone = 7 in. and its height = 29 in.

Base area of the cone = πr2
[Formula.]

= 3.14 × 72
[Substitute the values.]

= 3.14 × 49
[Substitute the value of 72 = 49.]

= 153.86
[Multiply.]

The base area of the cone = 153.86 in.2

Correct answer : (1)
20.
Radius and the height of a right circular cone are $\frac{12}{\pi }$ in. and $\frac{16}{\pi }$ in.. What is the curved surface area of the right circular cone? (Round the answer to the nearest hundredth of a unit.) a. 276.43 sq. in. b. 176.43 sq. in. c. 126.43 sq. in. d. 76.43 sq. in.

#### Solution:

Radius of the right circular cone = 12 / πin.

Height of the cone = 16 / π in.

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

= (16π)2+(12π)2
[Substitute the values.]

= 256π2+144π2
[Substitute the square values.]

=144+256π2
[Add fractions.]

= 400π2
[Add 144 and 256.]

= 20 / π
[Simplify.]

Slant height of the cone = 20 / π in.

Curved surface area of the right circular cone = πrl
[Formula.]

= π × 20π × 12π
[Substitute the values.]

= 240π
[Multiply.]

= 240 / 3.14
[Substitute the value of π = 3.14.]

= 76.4331
[Simplify.]

76.43 sq.in.
[Round to the nearest hundredths.]

Correct answer : (4)

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