# Surface Area of Pyramids and Cones Worksheet

Surface Area of Pyramids and Cones Worksheet
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1.
What is the base area of the square pyramid whose lateral edge is 5 cm. and the height is 4 cm.?

 a. 9 cm2 b. 18 cm2 c. 36 cm2 d. 10 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 + 42 = 52
[Substitute height, AB = 4 and lateral edge, AC = 5.]

BC2 + 16 = 25
[Simplify.]

BC2 = 9
[Subtract 16 from both sides.]

BC = 9 = 3 cm.
[Take square root on both sides.]

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

= 2 × 3 = 6 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.]

= 1 / 2 × 6 × 6 = 18
[The diagonals of a square have equal measures. Substitute the values.]

The base area of the square pyramid is 18 cm.2.

2.
The circumference of the base of a conical tent is 50.24 m and its slant height is 12 m. Find the area of the canvas used in making the tent.
 a. 298.44 m2 b. 304.44 m2 c. 301.44 m2 d. 307.44 m2

#### Solution:

Area of canvas required = lateral area of the conical tent

Circumference of the conical tent = 2πr = 50.24 m

2 × π × r = 50.24

r = 50.242π

Lateral area of the tent = π rl
[Formula.]

= π × 50.242π × 12
[Substitute the values.]

= 301.44 m2
[Multiply.]

So, area of the canvas required is 301.44 m2.

3.
Find the slant height of the cone if its lateral area is 516 ft2 and its radius is 12 ft. (Round the answer to the nearest hundredth) Use π = 3.14.
 a. 135.02 ft b. 15.69 ft c. 13.69 ft d. 11.69 ft

#### Solution:

Lateral area of a cone = πrl = 516 ft2

Radius of the cone = 12 ft

π × 12 × l = 516
[Substitute the value of r in step 1.]

π × 12 × l12 = 51612
[Divide each side by 12.]

πl = 43
[Simplify.]

3.14 × l = 43
[Substitute π = 3.14.]

3.14 × l3.14 = 433.14
[Divide each side 3.14.]

l = 13.694 13.69
[Simplify.]

The slant height of the cone is 13.69 ft.

4.
The height(h) of a cone is 40 cm. and its diameter(d) is 60 cm. What is its slant height?
 a. 70 cm b. 60 cm c. 80 cm d. 50 cm

#### Solution:

Height of a cone, h = 40 cm and its diameter = 60 cm

Radius of the cone, r = diameter / 2 = 60 / 2 = 30 cm

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.]

= 402+302
[Substitute the values.]

= 1600 + 900
[Substitute the values of 402, 302.]

= 2500

= 50 cm
[Simplify.]

The slant height of the cone = 50 cm

5.
The curved surface area of a cone is 2307.90 mm2 and the radius of the base of the cone is 21 mm. What is its height? Use π = 3.14.
 a. 23 mm b. 28 mm c. 30 mm d. 33 mm

#### Solution:

The curved surface area of a cone = πrl = 2307.90 mm2 and its radius is 21 mm.

Substitute the r value in πrl = 2307.90 mm2

π × 21 × l = 2307.90

π × 21 × l21 = 2307.9021
[Divide each side by 21.]

π × l = 109.90
[Simplify.]

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

3.14 × l3.14 = 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 on both sides.]

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

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

784 = h2
[Simplify.]

784 =h2
[Square on both the sides.]

28 = h
[Simplify.]

The height of the cone = 28 mm

6.
The total surface area of a cone is 60 square inches. If its slant height is four times the radius, then what is the base diameter of the cone? Use $\pi$ = 3.
 a. 12 in. b. 8 in. c. 4 in. d. 2 in.

#### Solution:

The total surface area of a cone = πrl + πr2 = 60 in.2

Slant height l = 4 × radius = 4r

3 × r × 4 r + 3 × r2 = 60
[Substitute l = 4r and π = 3.]

12r2 + 3r2 = 60
[Multiply.]

15r2 = 60

15r215 = 60 / 15
[Divide each side by 15.]

r2 = 4
[Simplify.]

r = 4
[Take square root on both sides.]

r = 2
[Simplify.]

Base radius of the cone = 2 in.

Base diameter of the cone = 2 × radius = 2 × 2 = 4 in.

7.
A circular cone with a base radius of 13 cm has a surface area of 2122.64 cm2. What would be its slant height?
 a. 36 cm b. 39 cm c. 26 cm d. 52 cm

#### Solution:

Surface area of a cone = πrl + πr2 = 2122.64 cm2

π × 13 × l + π × 132 = 2122.64
[Substitute the value of the radius.]

13πl + 169π = 2122.64
[Multiply.]

13πl + 169π - 169π = 2122.64 - 169π
[Subtract 169π from each side.]

13πl = 2122.64 - 169π
[Simplify.]

l = 2122.64 - 169π13π = 2122.6413π - 13
[Divide each side by 13π.]

l = 39

The slant height of the cone is 39 cm.

8.
Find the surface area of the square pyramid. [Given $a$ = 12 cm, $b$ = 14 cm.]

 a. 580 cm2 b. 480 cm2 c. 500 cm2 d. 530 cm2

#### Solution:

From the figure, side of a square is 12 cm

= 12 × 12 × 14 = 84 cm2
Area of the lateral face = 1 / 2× base × height
[Lateral face is a triangle with base 12 cm and height 14 cm.]

Surface area of the square pyramid = base area + lateral area
[Formula.]

= base area + 4 × area of lateral face

= 12 × 12 + 4× 84
[Substitute the values.]

= 144 + 336
[Multiply.]

= 480

The surface area of the square pyramid is 480 cm2.

9.
What is the slant height of the square pyramid if its height ($h$) is 8 cm and its base ($a$) is 12 cm?

 a. 60 cm b. 20 cm c. 10 cm d. 30 cm

#### Solution:

Height of the pyramid = 8 cm

Base of the pyramid is a square.

Length of each side of a square = 12 cm

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 = AB2 = 6 cm

EG2 = EF2 + FG2

EG2 = 82 + 62
[Substitute the values.]

EG2 = 82+62
[Take square root on both sides.]

The slant height of the pyramid with square base = 10 cm

10.
Christina distributed paper hats to children on a Christmas eve. The hats are in the shape of a cone with base radius of 9 cm and a slant height of 18 cm. Find the surface area of the paper used to make a hat. Use $\pi$ = 3.14.
 a. 513.98 cm2 b. 30.14 cm2 c. 508.68 cm2 d. 162 cm2

#### Solution:

Base radius of the hat, which is in the shape of cone, r = 9 cm and its slant height, l = 18 cm

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

= π × 9 × 18
[Substitute the values.]

= 162π
[Multiply.]

= 162 × 3.14
[Substitute the value of π = 3.14.]

= 508.68 cm2

So, the surface area of the paper used to make a hat = 508.68 cm2