Surface Area of Pyramids and Cones Worksheet - Page 5

Surface Area of Pyramids and Cones Worksheet
• Page 5
41.
The curved surface area of a cone is 292.02 cm.2 and its slant height is 93 cm. Find the total surface area of the cone. (Round the answer to the nearest hundredth unit.) [Use π = 3.14.]
 a. 299.16 cm2 b. 301.56 cm2 c. 295.16 cm2 d. 291.16 cm2

Solution:

Curved surface area of cone = πrl
[Formula.]

Curved surface area of a cone = 292.02 cm.2 and its slant height = 93 cm.

π rl = 292.02
[From the steps 1 and 2.]

π × r × 93 = 292.02
[Substitute the value l = 93.]

93πr93= 292.02 / 93
[Divide each side by 93.]

πr = 3.14

3.14 × r = 3.14
[Substitute π = 3.14.]

3.14r3.14 = 3.143.14
[Divide each side by 3.14.]

r = 1 cm

Total surface area of the cone = πrl + πr2
[Formula.]

= 292.02 + 3.14 × 12
[Substitute the value of π and curved surface area.]

= 292.02 + 3.14 × 1

= 292.02 + 3.14
[Simplify.]

= 295.16

The total surface area of the cone = 295.16 cm2.

42.
What length of canvas, 4 m. in width is required to make a conical tent 14 m. in diameter and 6.3 m. slant height? Also find the cost of the canvas at the rate of $13 per meter. (Round the answer to the nearest hundredth of a unit.) [Use π = 3.14.]  a. Length = 30.62 m, cost =$446.06 b. Length = 38.62 m, cost = $454.06 c. Length = 34.62 m, cost =$454.06 d. Length = 34.62 m, cost = $450.06 Solution: Diameter of the conical tent = 14 m. and its slant height, l = 6.3 m. Radius of the tent, r = diameter / 2 = 14 / 2 = 7 m. Lateral area of the tent = πrl [Formula.] = 3.14 x 7 x 6.3 [Substitute the values.] = 138.474 m.2 [Multiply.] Lateral area of the tent = 138.474 m.2 Required canvas = lateral area of the tent = 138.474 m.2 Width of the canvas = 4 m. Length of the canvas = area of canvas/width of canvas = 138.4744 [Substitute the values.] = 34.618 m 34.62 m Length of the canvas = 34.62 m. Cost of the canvas =$13 per meter.

Total cost of required canvas = 34.62 x 13 = $450.06 The length of the canvas is 34.62 m. and it's total cost is$450.06

43.
A cone made of paper is 24 in. high and has a radius of 7 in. Find the area of the paper needed to construct the cone. [Use π = 3.14.]

 a. 681.38 in.2 b. 703.36 in.2 c. 527.52 in.2 d. 549.5 in.2

Solution:

Height of the cone, h = 24 in. and its base radius, r = 7 in.

Slant height of the tent, l = h2 +r2
[Formula.]

= 242 +72
[Substitute the values.]

= (576 + 49)
[Substitute the values of 242 and 72.]

= 625

= 25
[Simplify.]

Slant height of the cone, l = 25 in.

Surface area of the cone = πrl + πr2
[Formula.]

= (3.14 × 7 × 25) + (3.14 × 7 × 7)
[Substitute the values.]

= 549.5 + 153.86 = 703.36
[Simplify.]

Therefore, 703.36 in.2 paper is needed to construct the cone.

44.
You need to paint the outside of the house as shown in the figure. A gallon of paint covers about 400 square feet. How many gallons will you need to buy? (Ignore the areas of window and door.)

 a. 5 gallons b. 2.5 gallons c. 6 gallons d. 4.5 gallons

Solution:

From the figure, the bottom portion of the house represents a cube and the 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 the square in each face of the cube = 20 ft

Area of the square = side x side = 20 × 20 = 400 ft2

Base of the triangle = 20 ft and its height = 10 ft

Area of the triangle = 1 / 2 × base × height = 1 / 2 × 20 × 10 = 100 ft2

Out side area of the house = 4 × 400 + 4 × 100
[Substitute the values in the step 2.]

= 1600 + 400
[Multiply.]

= 2000

Out side area of the house = 2000 ft2

A gallon of paint covers the area = 400 ft2

Number of gallons required to cover the area of the out side of the house = 2000 / 400 = 5.

The number of gallons required to paint the out side of the house = 5.

45.
From the two figures shown, which has the larger surface area?

 a. Pyramid b. Both cannot be compared c. Both cone and pyramid have equal surface areas. d. Cone

Solution:

From the figure, slant height of the pyramid = slant height of the cone = 13 in.

Length of each side of the base of the pyramid = diameter of the cone = 10 in.

Radius of the cone = diameter of the cone/2 = 10 / 2 = 5 in.

Surface area of the cone = πrl + πr2
[Formula.]

= 3.14 x 5 x 13 + 3.14 x 52
[Substitute the values.]

= 204.1 + 78.5
[Multiply.]

= 282.6

Surface area of cone = 282.6 in.2

Surface area of the pyramid = base area + 4 x area of triangle
[Formula.]

= (side x side) + 4 x 1 / 2 x base x height

= (10 x 10) + 4 x 12 x 10 x 13
[Substitute the values.]

= 100 + 260
[Multiply.]

= 360

Surface area of the pyramid = 360 in.2

So, the pyramid has a larger surface area.

46.
What length of paper, of width 30 in., will be required to make a hollow cone of radius 12 in. and slant height 45 in.?
 a. 1696 in. b. 678.24 in. c. 56.52 in. d. 18 in.

Solution:

Radius of the cone = 12 in. and its slant height = 45 in.

Lateral area of the cone = πrl
[Formula.]

= 3.14 x 12 x 45
[Substitute the values.]

= 1695.60 in.2
[Multiply.]

The lateral area of the cone = 1695.60 in.2

Area of the required paper = lateral area of the cone = 1695.60 in.2

Width of the paper = 30 in.

Length of the paper = area of the paper/width of the paper
[Formula.]

= 1695.6030
[Substitute the values.]

= 56.52 in.
[Simplify.]

The length of the paper = 56.52 in.

47.
The slant height of a cone is 35 in. and its base radius is 28 in. What is its height?

 a. 27 in. b. 23 in. c. 21 in. d. 25 in.

Solution:

Slant height of a cone = 35 in. and its base radius = 28 in.

Let h = height of the cone.

Slant height = √ (h2 + r2)
[Formula.]

35 = √ (h2 + 282)
[Substitute the values.]

352 = (√ (h2 + 282))2
[Squaring both sides.]

1225 = h2 + 784
[Simplify.]

1225 - 784 = h2 + 784 - 784
[Subtract 784 from both sides.]

441 = h2
[Simplify.]

√441 = √h2
[Taking square root on both sides.]

21 = h
[Simplify.]

The height of the cone = 21 in.

48.
The curved surface area of a cone, whose slant height is 3 times the radius of its base, is 75$\pi$. Find the slant height of the cone.
 a. 15 units b. 13 units c. 19 units d. 17 units

Solution:

Curved surface area of a cone = πrl = 75π

Let r = base radius of the cone.

Slant height of the cone (l) = 3 x radius = 3r.

π x r x 3r = 75π
[Substitute the value of l = 3r in πrl = 75π.]

3πr2 = 75π
[Multiply.]

3πr2/3π = 75π
[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 x 5 = 15 units

The slant height of the cone = 15 units.

49.
What is the total surface area of the figure?

 a. 188.4 cm2 b. 160.14 cm2 c. 216.66 cm2 d. 244.92 cm2

Solution:

From the figure, the base radius = 3 cm., height of cylinder = 5 cm. and slant height of the cone = 7 cm.

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

= πr2 + 2πrh + πrl

= π x 32 + 2π x 3 x 5 + π x 3 x 7
[Substitute the values.]

= 9π + 30π + 21π
[Multiply.]

= 60π

= 60 x 3.14
[Substitute the value of π =3.14.]

= 188.4 cm.2
[Multiply.]

The surface area of the figure = 188.4 cm.2.

50.
What is the expression for the surface area of a cone, whose slant height is 4 times the radius of its base?
 a. 5$\pi$$r$2 unit2 b. 7$\pi$$r$2 unit2 c. 6$\pi$$r$2 unit2 d. 4$\pi$$r$2 unit2

Solution:

Let r = base radius of a cone.

Slant height of the cone = 4 times the radius = 4 x r = 4r units

Surface area of the cone = πrl + πr2
[Formula.]

= π x r x 4r + πr2
[Substitute l = 4r.]

= 4πr2 + πr2
[Multiply.]

= 5πr2