Surface Area of Prisms and Cylinders Worksheet

Surface Area of Prisms and Cylinders Worksheet
• Page 1
1.
Which of the following triangles is similar to the ΔPQR?

 a. ΔQRS b. ΔPNS c. ΔSQM d. ΔSRN

Solution:

Similar figures have same shape but not necessarily of same size.

ΔPQR is a right triangle.
[From the figure.]

ΔPNS, ΔSRN, and ΔSQM are not similar to ΔPQR as they are not right triangles.

From the given choices the triangle that is similar to ΔPQR is ΔQRS.

2.
What is the surface area of the cylinder shown?

 a. 234.68 cm2 b. 192.58 cm2 c. 220 cm d. 212.14 cm2

3.
The diameter of a roller is 110 cm and its length is 150 cm. It takes 800 complete revolutions to move once over a play ground to level it. Find the area of the play ground. [Take $\pi$ = 3.]
 a. 1980 m² b. 4686 m² c. 3960 m² d. 1320 m²

Solution:

Area of the play ground = No. of revolutions × Curved surface area of the rollers.
[Formula.]

Radius of the roller = d / 2= 55 cm
Length of the roller = 150cm

Curved surface area = 2 × 3 × 55 × 150 = 49500 cm2
[Curved surface area = 2 π r l.]

Area of the play ground = 800×49500 / 10000= 3960 m2
[Substitute in step1 and simplify.]
[1m = 100cm.]

4.
The inner diameter of a circular well is 7 m and it is 6 m deep. Find the cost of plastering its inner curved surface at the rate of $25 per sq m. [Take $\pi$ = 3.]  a.$1565 b. $1575 c.$1595 d. $1615 Solution: Cost of plastering = (Rate of plastering) × (Inner curved surface area). [Formula.] Inner radius = d / 2 = 3.50 m [Radius = Diamater / 2.] Depth = 6 m [Given.] Inner surface area = 3 × 3.50 × 6 = 63 m2 [Well is in the form of a cylinder.] [Inner surface area of a cylinder = π r h.] Rate of plastering per sq.m =$25
[Given.]

Cost of plastering = 25 × 63 = $1575 [Substitute in step1 and simplify.] Correct answer : (2) 5. A closed circular water tank of outer diameter 4 m and height 2 m is to be painted. Find the total charges for painting if the rate of painting is$10 per sq.m. [Take $\pi$ = 3.]
 a. $623 b.$603 c. $437 d.$480

Solution:

Cost of painting = (Rate of painting) × (total surface area of the tank).
[Formula.]

Total surface area = Curved surface area + area of the end faces.
[Analysis.]

Radius = Diameter / 2 = 2 m

Height = 2 m.
[Given.]

Curved surface area = 2 × 3 × 2 × 2 = 24 sq.m
[Curved surface area = 2 π r h.]

Area of the end faces = 2 × 3 × 22 = 24 sq.m
[Area of end faces = 2 π r2.]

Total surface area = 24 + 24 = 48 sq.m
[Substitute in step2.]

Rate of painting = $10 per sq.m [Given.] Cost of painting = 10 × 48 =$480
[Substitute in step1 and simplify.]

6.
A cylindrical open vessel is to be made of a thin sheet of uniform thickness. If the diameter of the vessel is 30 cm and the height is 22 cm, then find the area of the sheet required to make it. [Take $\pi$ = 3.]
 a. 2778 cm2 b. 2582 cm2 c. 2753 cm2 d. 2655 cm2

Solution:

Area of the sheet = Curved surface area of the cylinder + Base area of the vessel.
[Analysis.]

Radius of the vessel = d / 2 = 15 cm
[Given.]

Height of the vessel = 22 cm
[Given.]

Curved surface area = 2πrh = 2 × 3 × 15 × 22 = 1980 sq.cm
[Formula.]

Area of the base = π × r2 = 3 × 152 = 675 sq.cm
[Formula.]

Area of the sheet = 1980 + 675 = 2655 cm2
[From steps 4 and 5.]

7.
The inner and outer diameter of a metallic pipe are 7 cm and 8 cm. Length of the pipe is 4 m. Find the total surface area of the pipe in sq.cm. [Take $\pi$ = 3.]
 a. 18022.50 sq.cm b. 17946.50 sq.cm c. 18089.50 sq.cm d. 18198.50 sq.cm

Solution:

Total surface area of the pipe = Inner curved surface area + Outer curved surface area + Area of two end faces.
[Formula.]

Inner curved surface area = 2 × 3 × 3.5 × 4 × 100 = 8400 sq.cm
[Inner curved surface area = 2 π a h.]

Outer curved surface area = 2 × 3 × 4 × 4 × 100 = 9600 sq.cm
[Outer curved surface area = 2 π b h.]

Area of the end faces = 2(3 × 42 - 3 × 3.52) = 22.50 sq.cm
[Area of the circular ring = 2(π b2 - π a2).]

Total surface area = 8400 + 9600 + 22.50 = 18022.50 sq.cm
[Substitute in step1 and simplify.]

So, the total surface area of the pipe is 18022.50 sq.cm

8.
Find the surface area of the cylinder.

 a. 16$\pi$ b. 128$\pi$ c. 52$\pi$ d. 96$\pi$

9.
Find the area of the aluminium sheet required to make a cylindrical can measuring a base diameter of 8 cm. and a height of 12 cm. Round the answer to the nearest ten.
 a. 401 cm2 b. 401.92 cm2 c. 402.92 cm2 d. 402 cm2

Solution:

The surfaces of a cylinder include two circles and one rectangle.

Therefore, the total surface area = 2 × area of the circle + area of the rectangle.

= 2 π r2 + b h

= 2 π (4 × 4) + 2 π (4 × 12)
[Diameter = radius / 2and substituting.]

= π (32 + 96) = π (128) = 401.92 cm.2
[Simplify.]

The area of the aluminium sheet needed to make the cylindrical can is 401.92 cm.2 and to the nearest ten is 402 cm.2.

10.
The figure shows a rectangular prism with a cylinder which has been removed from it. What is the total surface area of the given solid? Round the answer to the nearest ten.

 a. 263 m2 b. 232 m2 c. 231 m2 d. 230 m2

Solution:

Total surface area of the rectangular prism = surface area of the rectangular prism + lateral area of the cylinder - 2 × base area of the cylinder.

Dimensions of the rectangular prism are 4m. × 6m. × 8m.

Therefore, the surface area of the rectangular prism = 2 (lw + wh + lh)

= 2 (8 × 4 + 4 × 6 + 6 × 8) = 208
[Substituting the values.]

Lateral area of the cylinder = 2 π r h
[Formulae.]

2 (π × 2.5 × 4) = 20 π = 62.8
[Substituting the values and simplify.]

Base area of the cylinder = π r2
[Formulae.]

= π (2.5 × 2.5) = 6.25 π = 19.625
[Substituting the values.]

The total surface area of the rectangular prism = 208 + 62.8 - 2 × 19.625

= 208 + 62.8 - 39.25 = 231.55 m.2
[Simplify.]

231.55 to the nearest ten is 232 m.2.

Therefore, the surface area of the solid to the nearest ten is 232 m.2.