﻿ Surface Area of Pyramids and Cones Worksheet - Page 4 | Problems & Solutions
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Surface Area of Pyramids and Cones Worksheet
• Page 4
31.
What is the slant height of the pyramid, if $h$ is the height of a pyramid standing on a square base of side $a$ units?  a. √ (4 $h$2 + $a$2)/2 units b. √ (2 $h$2 + $a$2)/2 units c. (2 $h$2 + $a$2) units d. √ (4 $h$2 + 2 $a$2)/2 units

#### Solution:

Height of the pyramid = h units

Base of the pyramid is a square.

Length of each side of a square = a units.

Slant height of the pyramid = EG

From the figure, EFG is a right triangle.

According to the Pythagoras theorem EG2 = EF2 + FG2
[Formula.]

FG is parallel to AB and length of FG = AB / 2 = a / 2 units.

EG2 = EF2 + FG2

EG2 = h2 + ( a / 2)2
[Substitute the values.]

EG2 = h2 + a2/4
[Simplify (a/22).]

EG2 = (4 h2 + a2)/4
[Add.]

√EG2 = √ ((4 h2 + a2)/4)
[Take square root on both sides.]

EG = √ (4h2 + a2)/2
[Simplify.]

The slant height of the pyramid with a square base = √ (4 h2 + a2)/2 units.

Correct answer : (1)
32.
Perpendicular distance from the base to the opposite vertex is called __________. a. Base b. Either height or slant height c. Slant height d. Height

#### Solution:

The perpendicular distance from the base to the opposite vertex is called the height.

Correct answer : (4)
33.
Perpendicular distance from the edge of the base to the opposite vertex of a pyramid is called _________________. a. Slant height b. Height c. Base d. None of the above

#### Solution:

Perpendicular distance from the edge of the base to the opposite vertex is called the slant height.

Correct answer : (1)
34.
Laura distributed paper hats to children on the Christmas eve. The hats are in the shape of a cone with a base radius of 6 cm. and a slant height of 17 cm. Find the surface area of the paper used to make each hat. [Use π = 3.14.] a. 102 cm2 b. 160.14 cm2 c. 320.28 cm2 d. 336.58 cm2

#### Solution:

Base radius of the hat, which is in the shape of cone, r = 6 cm. and its slant height, l = 17 cm.

Curved surface area of the hat = πrl
[Formula.]

= π × 6 × 17
[Substitute the values.]

= 102π
[Multiply.]

= 102 × 3.14
[Substitute the value of π = 3.14.]

= 320.28 cm.2

So, the surface area of the paper used to make a hat = 320.28 cm.2

Correct answer : (3)
35.
A circular cone is 6 in. high. The radius of the base is 8 in. What is the curved surface area of the cone? a. 251.2 in.2 b. 257.2 in.2 c. 259.8 in.2 d. 245.2 in.2

#### Solution:

Height of a circular cone, h = 6 in. and its base radius, r = 8 in.

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

= 62 + 82
[Substitute the values.]

= (36 + 64)
[Substitute the values of 62 and 82.]

= 100
[Add.]

= 10
[Simplify.]

Slant height of the cone = 10 in.

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

= 3.14 × 8 × 10
[Substitute the values.]

= 251.2 in.2

The curved surface area of the cone is 251.2 in.2.

Correct answer : (1)
36.
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. 65.08 cm2 b. 75.48 cm2 c. 69.08 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 is 69.08 cm.2.

Correct answer : (3)
37.
The circumference of the base of a conical tent is 25.12 m. and its slant height is 8 m. Find the area of the canvas used in making the tent.[Use $\pi$ = 3.14.] a. 102.48 m2 b. 106.48 m2 c. 98.48 m2 d. 100.48 m2

#### Solution:

Area of canvas required = lateral area of the conical tent

Circumference of the conical tent = 2πr = 25.12 m.

2 x 3.14 x r = 25.12
[Substitute the value of π = 3.14.]

6.28r = 25.12
[Multiply 2 with 3.14.]

6.28r6.28 = 25.126.28
[Divide each side by 6.28.]

r = 4
[Simplify.]

Lateral area of the tent = πrl
[Formula.]

= 3.14 x 4 x 8
[Substitute the values.]

= 100.48 m.2
[Multiply.]

So, area of the canvas required is 100.48 m.2.

Correct answer : (4)
38.
What is the slant height of the cone, if its lateral area is 720 ft2 and its radius is 12 ft? (Round the answer to the nearest hundredth) [Use π = 3.14.]  a. 17.11 ft b. 19.11 ft c. 21.11 ft d. None of the above

#### Solution:

Lateral area of a cone = πrl = 720 ft2

Radius of the cone = 12 ft

π x 12 x l = 720
[Substitute the value of r in step 1.]

(π x 12 x l)/12 = 72012
[Divide each side by 12.]

πl = 60
[Simplify.]

3.14 x l = 60
[Substitute π = 3.14.]

(3.14 x l)/3.14 = 603.14
[Divide each side 3.14.]

l = 19.108 19.11
[Simplify.]

The slant height of the cone is 19.11 ft.

Correct answer : (2)
39.
The height of a cone is 16 in. and its diameter is 24 in. What is the slant height of the cone?  a. 22 in. b. 16 in. c. 18 in. d. 20 in.

#### Solution:

Height of a cone, h = 16 in. and its diameter = 24 in.

Radius of the cone, r = diameter / 2 = 24 / 2 = 12 in.

The radius of the cone, height and the slant height form a right triangle.

So, slant height of the cone = √(h2 + r2)
[Apply Pythagorean theorem.]

= √(162 + 122)
[Substitute the values.]

= √(256 + 144)
[Substitute the values of 162, 122.]

= √400
[Add.]

= 20 in.
[Simplify.]

The slant height of the cone = 20 in.

Correct answer : (4)
40.
The curved surface area of a cone is 2307.90 mm.2 and the radius of the base of the cone is 21 mm. What is its height? [Use π = 3.14.] a. 30 mm b. 33 mm c. 28 mm d. 23 mm

#### Solution:

The curved surface area of a cone = πrl = 2307.90 mm.2 and its radius is 21 mm.

Substitute the r value in πrl = 2307.90 mm.2

π x 21 x l = 2307.90

(π x 21 x l)/21 = 2307.9021
[Divide each side by 21.]

π x l = 109.90
[Simplify.]

3.14 x l = 109.90
[Substitute π = 3.14.]

(3.14 x l)/3.14 = 109.903.14
[Divide each side by 3.14.]

l = 35 mm
[Simplify.]

Slant height of the cone = 35 mm.

Let h be the height of the cone.

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

35 = √(h2 + 212)
[Substitute the values.]

352 = (√(h2 + 212))2
[Squaring both sides.]

1225 = h2 + 441
[Substitute the values of 352 and 212.]

1225 - 441 = h2 + 441 - 441
[Subtract 441 from each side.]

784 = h2
[Simplify.]

√784 = √h2
[Square on both the sides.]

28 = h2
[Simplify.]

The height of the cone = 28 mm.

Correct answer : (3)

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