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Surface Area of Pyramids and Cones Worksheet - Page 6

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 6 π cm. and 8 π 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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