﻿ Surface Area of Pyramids and Cones Worksheet - Page 6 | Problems & Solutions
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Surface Area of Pyramids and Cones Worksheet
• Page 6
51.
What is the base area of a cone with a radius of 8 in. and a height of 28 in.? a. 190.09 in.2 b. 200.96 in.2 c. 211.83 in.2 d. 221.83 in.2

#### Solution:

Radius of a cone = 8 in. and its height = 28 in.

Base area of the cone = πr2
[Formula.]

= 3.14 × 82
[Substitute the values.]

= 3.14 × 64
[Substitute the value of 82 = 64.]

= 200.96
[Multiply.]

The base area of the cone is 200.96 in.2.

Correct answer : (2)
52.
The radius and the height of a right circular cone are $\frac{6}{\pi }$ cm. and $\frac{8}{\pi }$ cm. respectively. What is the curved surface area of the right circular cone? (Round the answer to the nearest hundredth of a unit.) a. 23.21 cm b. 19.11 cm c. 21.61 cm d. 15.11 cm

#### Solution:

Radius of the right circular cone = 6 / π cm.

Height of the cone = 8 / π cm.

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

= √ (( 8π )2 + ( 6π )2)
[Substitute the values.]

= √ (64/π2 + 36/π2)
[Substitute the square values.]

= √ ((64 + 36)/π2)
[Add fractions.]

= √ (100/π2)
[Add 64 and 36.]

= 10π
[Simplify.]

Slant height of the cone = 10 / π cm.

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

= π x 6π x 10π
[Substitute the values.]

= 60π
[Multiply.]

= 603.14
[Substitute the value of π = 3.14.]

= 19.1082
[Simplify.]

19.11
[Round to the nearest hundredths.]

Correct answer : (2)
53.
The base area of a cone is 452.16 in.2 and its height is 16 in. What is the surface area of the cone? a. 1216.21 in.2 b. 1205.76 in.2 c. 1226.41 in.2 d. 1195.31 in.2

#### Solution:

Base area of a cone = πr2 = 452.16 in.2 and its height = 16 in.

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 the square root on both sides.]

r = 12
[Simplify.]

[Formula.]

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

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

= 400
[Add.]

= 20 in.
[Simplify.]

Slant height of the cone = 20 in.

Surface area of the cone = πrl + πr2

= 3.14 × 12 × 20 + 3.14 × 122

= 753.60 + 452.16

= 1205.76

The surface area of the cone is 1205.76 in.2.

Correct answer : (2)
54.
What is the surface area of the figure?  a. 742.78 mm2 b. 723.56 mm2 c. 769.3 mm2 d. 722.78 mm2

#### Solution:

From the figure, the base diameter of the cylinder = 10 mm., the height of the cylinder = 15 mm. and the slant height of the cone = 14 mm.

Base radius of the cylinder = diameter / 2 = 10 / 2 = 5 mm.

Surface area of the figure = base area + lateral area of the cylinder + lateral area of the cone
[Formula.]

= πr2 + 2πrh + πrl

= π x 52 + 2π x 5 x 15 + π x 5 x 14
[Substitute the values.]

= 25π + 150π + 70π
[Multiply.]

= 245π
[Add.]

= 245 x 3.14
[Substitute the π= 3.14.]

= 769.3 mm.2
[Multiply.]

The surface area of the figure is 769.3 mm.2.

Correct answer : (3)
55.
What is the base area of a square pyramid with lateral edge($a$) 15 cm and height($h$) 12 cm?  a. 81 cm2 b. 18 cm2 c. 324 cm2 d. 162 cm2

#### Solution:

ΔABC is a right triangle.
[Height AB is perpendicular to BC and the lateral edge AC is the hypotenuse.]

BC2 + AB2 = AC2
[Apply Pythagorean theorem.]

BC2 + 122 = 152
[Substitute height, AB = 12 and lateral edge, AC = 15.]

BC2 + 144 = 225
[Simplify.]

BC2 = 81
[Subtract 144 from both sides.]

BC = 81 = 9 cm
[Take square root on both sides.]

Diagonal of the square base = 2 × BC
[BC is half the diagonal of the square base.]

= 2 × 9 = 18 cm
[Substitute and multiply.]

Area of a square = 1 / 2 × product of the diagonals
[Formula for the area of a square in terms of the measure of diagonals.]

= 12 × 18 × 18 = 162
[The diagonals of a square have equal measures. Substitute the values.]

The base area of the square pyramid is 162 cm2.

Correct answer : (4)
56.
What is the base area of the square pyramid whose lateral edge(b) is 8 cm and the height(a) is 7 cm?  a. 80 cm2 b. 130 cm2 c. 30 cm2 d. 50 cm2

#### Solution:

ΔABC is a right triangle.
[Height AB is perpendicular to BC and the lateral edge AC is the hypotenuse.]

BC2 + AB2 = AC2
[Apply Pythagorean theorem.]

BC2 + 72 = 82
[Substitute height, AB = 7 and lateral edge, AC = 8.]

BC2 + 49 = 64
[Simplify.]

BC2 = 15
[Subtract 49 from both sides.]

BC = 15
[Take square root on both sides.]

Diagonal of the square base = 2 × BC
[BC is half the diagonal of the square base.]

= 2 × 15
[Substitute and multiply.]

Area of a square = 1 / 2 × product of the diagonals
[Formula for the area of a square in terms of the measure of diagonals.]

= 1 / 2 × 215 × 215 = 30 cm2
[The diagonals of a square have equal measures. Substitute the values.]

The base area of the square pyramid is 30 cm2.

Correct answer : (3)
57.
Find the volume of the rectangular pyramid.  a. 40 cm3 b. 100 cm3 c. 160 cm3 d. 140 cm3

#### Solution:

From the figure, the length, width and height of the rectangular pyramid are 5 cm, 4 cm, and 6 cm, respectively.

Volume of a rectangular pyramid = (1 / 3) × length × width × height

= (13) × 5 × 4 × 6
[Substitute the values.]

= 1203 cm3

= 40 cm3
[Simplify.]

The volume of the rectangular pyramid is 40 cm3.

Correct answer : (1)

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