# DAV Class 8 Maths Chapter 14 Worksheet 4 Solutions

The DAV Maths Book Class 8 Solutions Pdf and DAV Class 8 Maths Chapter 14 Worksheet 4 Solutions of Mensuration offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 14 WS 4 Solutions

Question 1.
Find the curved surface area and total surface area of a right circular cylinder whose height is 15 cm and the radius of the base is 7 cm.
Solution:
Here r = 7 cm and h = 15 cm
∴ Curved surface area = 2πrh
= 2 × $$\frac{22}{7}$$ × 7 × 15
= 660 cm2
Total surface area = 2πrh + 2πr2
= 2πr[h + r]
= 2 × $$\frac{22}{7}$$ × 7 [15 + 7]
= 44 × 22
= 968 cm2 Question 2.
The diameter of the base of a right circular cylinder is 4.2 dm and its height is 1 dm. Find the area of the curved surface in cm2.
Solution:
Here r = $$\frac{4.2}{2}$$
= 2.1 dm
= 21 cm
h = 1 dm = 10 cm
∴ Curved surface area = 2πrh
= 2 × $$\frac{22}{7}$$ × 21 × 10
= 1320 cm2

Question 3.
The curved surface area of a cylinder is 1320 cm2 and its base has a radius of 10.5 cm. Find the height of the cylinder.
Solution:
Curved surface area = 2πrh
⇒ 1320 = 2 × $$\frac{22}{7}$$ × 10.5 × h
⇒ $$\frac{1320 \times 7}{2 \times 22 \times 10.5}$$
⇒ h = 20 cm

Question 4.
The circumference of the base of a cylinder is 176 cm and its height is 65 cm. Find its lateral surface area.
Solution:
Here 2πr = 176 cm and h = 65 cm
∴ Lateral surface area = 2πrh
= 176 × 65
= 11440 cm2 Question 5.
The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their curved surface areas.
Solution:
Let the radii of the two cylinders be 2x cm and 3x cm and their respective heights be 5y cm and 3y cm.
∴ C.S.A. of 1st cylinder = 2πr1h1
∴ C.S.A. of 2nd cylinder = 2πr2h2
∴ $$\frac{\text { C.S.A. of } 1 \text { st cylinder }}{\text { C.S.A. of } 2 \text { nd cylinder }}=\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}=\frac{r_1 h_1}{r_2 h_2}=\frac{2 x \times 5 y}{3 x \times 3 y}=\frac{10}{9}$$
Hence, the required ratio = 10 : 9.

Question 6.
The outer diameter of a gas cylinder is 20 cm. Find the cost of painting its outer curved surface at ₹ 4 per square centimetre if the height of the cylinder is 70 cm.
Solution:
The curved surface area of the cylinder = 2πrh
= 2 × $$\frac{22}{7}$$ × 10 × 70
= 4400 cm2
∴ Cost of painting = 4 × 4400 = ₹ 17600.

Question 7.
A cylindrical vessel, open at the top, has a radius of 10 cm and a height of 14 cm. Find the total surface area of the vessel. (Take π = 3.14)
Solution:
∴ T.S.A. of the cylindrical vessel = 2πrh + πr2
= πr(2h + r)
= 3.14 × 10 (2 × 14 + 10)
= 31.4 × 38
= 1193.2 cm2

Question 8.
The length of a roller is 40 cm and its diameter is 21 cm. It takes 300 complete revolutions to move once over to level the floor of a room. Find the area of the room in m2.
Solution:
Here h = 40 cm, r = $$\frac{21}{2}$$ cm
∴ Curved surface area = 2πrh
= 2 × $$\frac{22}{7}$$ × $$\frac{21}{2}$$ × 40
= 2640 cm2
= 0.2640 m2 Question 9.
A closed metallic cylinder has a base diameter 56 cm and it is 2.25 m high. Find the cost of the metal used to make it if it costs ₹ 80 per sq. m.
Solution:
Here r = $$\frac{56}{2}$$ = 28 cm
h = 2.25 m = 225 cm
∴ T.S.A. = 2πrh + 2πr2
= 2πr[h + r]
= 2 × $$\frac{22}{7}$$ × 28 × [225 + 28]
= 176 × 253
= 44528 cm2
= 4.4528 m2
∴ The cost of metal = ₹ 80 × 4.4528 = ₹ 358.25.

Question 10.
A rectangular piece of paper is 22 cm in length and 10 cm in width. It is rolled into a cylinder along its length. Find the surface area of the cylinder.
Solution:
Here 2πr = 22 and h = 10 cm
∴ The curved surface area of the cylinder so formed = 2πrh
= 22 × 10
= 220 cm2 Question 11.
A cylindrical well is 21 m deep. Its outer and inner diameters are 21 m and 14 m respectively. Find the cost of renovating the inner curved surface and outer curved surface at the rate of ₹ 25 per m2.
Solution:
The curved surface area of the inner and outer sides = 2πr1h1 + 2πr2h2
= 2πh[r1 + r2]
= 2 × $$\frac{22}{7}$$ × 21 × [$$\frac{21}{2}$$ + 7]
= 2 × $$\frac{22}{7}$$ × 21 × $$\frac{35}{2}$$
= 2310 m2
∴ Cost of renovation = ₹ 25 × 2310 = ₹ 57750.