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 cm^{2}

Total surface area = 2πrh + 2πr^{2}

= 2πr[h + r]

= 2 × \(\frac{22}{7}\) × 7 [15 + 7]

= 44 × 22

= 968 cm^{2}

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 cm^{2}.

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 cm^{2}

Question 3.

The curved surface area of a cylinder is 1320 cm^{2} 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 cm^{2}

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 = 2πr_{1}h_{1}

∴ C.S.A. of 2nd cylinder = 2πr_{2}h_{2}

∴ \(\frac{\text { C.S.A. of } 1 \text { st cylinder }}{\text { C.S.A. of } 2 \text { nd cylinder }}=\frac{2 \pi r_1 h_1}{2 \pi r_2 h_2}=\frac{r_1 h_1}{r_2 h_2}=\frac{2 x \times 5 y}{3 x \times 3 y}=\frac{10}{9}\)

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 cm^{2}

∴ 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 + πr^{2}

= πr(2h + r)

= 3.14 × 10 (2 × 14 + 10)

= 31.4 × 38

= 1193.2 cm^{2}

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 m^{2}.

Solution:

Here h = 40 cm, r = \(\frac{21}{2}\) cm

∴ Curved surface area = 2πrh

= 2 × \(\frac{22}{7}\) × \(\frac{21}{2}\) × 40

= 2640 cm^{2}

= 0.2640 m^{2}

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 + 2πr^{2}

= 2πr[h + r]

= 2 × \(\frac{22}{7}\) × 28 × [225 + 28]

= 176 × 253

= 44528 cm^{2}

= 4.4528 m^{2}

∴ 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 cm^{2}

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 m^{2}.

Solution:

The curved surface area of the inner and outer sides = 2πr_{1}h_{1} + 2πr_{2}h_{2}

= 2πh[r_{1} + r_{2}]

= 2 × \(\frac{22}{7}\) × 21 × [\(\frac{21}{2}\) + 7]

= 2 × \(\frac{22}{7}\) × 21 × \(\frac{35}{2}\)

= 2310 m^{2}

∴ Cost of renovation = ₹ 25 × 2310 = ₹ 57750.