﻿ Surface Area of Pyramids and Cones Worksheet - Page 2 | Problems & Solutions

# Surface Area of Pyramids and Cones Worksheet - Page 2

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

= 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

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

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

= 69.08
[Multiply.]

The surface area of the cone = 69.08 cm2

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

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

= 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

= 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

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

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.

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.

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.

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.

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

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

= 400π2

= 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.]