In this chapter, we provide RD Sharma Solutions for Class 8 Chapter 22 Mensuration for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 8 Chapter 22 Mensuration pdf, free RD Sharma Solutions for Class 8 Chapter 22 Mensuration book pdf download. Now you will get step by step solution to each question.

### Access answers to RD Sharma Maths Solutions For Class 8 Exercise 22.1 Chapter 22 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)

**1. Find the curved surface area and total surface area of a cylinder, the diameter of whose base is 7 cm and height is 60 cm.**

**Solution:**

We have,

Diameter of cylinder = 7 cm

So, Radius of cylinder = 7/2 cm

Height of cylinder = 60 cm

By using the formula,

Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 7/2 × 60

= 1320 cm^{2}

Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7/2 (60 + 7/2)

= 22 (127/2)

= 1397 cm^{2}

**2. The curved surface area of a cylindrical road is 132 cm ^{2}. Find its length if the radius is 0.35 cm.**

**Solution:**

We have,

Curved surface area of cylindrical road =132 cm^{2}

Radius of road = 0.35 cm

Let length of road be ‘h’ cm

By using the formula,

Curved surface area of cylindrical road = 2πrh

So, 2πrh = 132

2 × 22/7 × 0.35 × h = 132

h = 132×7 / 2×22×0.35

= 924 / 15.4

= 60cm

∴ Length of road is 60 cm.

**3. The area of the base of a right circular cylinder is 616 cm ^{2} and its height is 2.5 cm. Find the curved surface area of the cylinder.**

**Solution:**

We have,

Area of base of right circular cylinder = 616 cm^{2}

Height of cylinder = 2.5 cm

Let the radius of cylinder be ‘r’ cm

By using the formula,

Area of base of right circular cylinder = πr^{2}

So, πr^{2} = 616

22/7 r^{2} = 616

r^{2} = 616×7 / 22

= 196

r = √196

= 14cm

∴ Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 14 × 2.5

= 1540/7

= 220 cm^{2}

**4. The circumference of the base of a cylinder is 88 cm and its height is 15 cm. Find its curved surface area and total surface area.**

**Solution:**

We have,

Circumference of base of cylinder = 88 cm

Height of cylinder = 15 cm

By using the formula,

Circumference of base of cylinder = 2πr

So,

2πr = 88

2 × 22/7 × r = 88

r = 88×7 / 44

= 616/44

= 14cm

Radius of cylinder = 14 cm

∴ Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 14 × 15

= 1320 cm^{2}

∴ Total surface area area of cylinder = 2πr (h+r)

= 2 × 22/7 × 14 (15 + 14)

= 2 × 22/7 × 14 × 29

= 2552 cm^{2}

**5. A rectangular strip 25 cm × 7 cm is rotated about the longer side. Find the total surface area of the solid thus generated.**

**Solution:**

We have,

Dimension of rectangular strip = 25 cm × 7cm

When this strip is rotated about its longer side,

Height of cylinder becomes = 25 cm

Radius = 7 cm

∴ Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7 (25 + 7)

= 2 × 22/7 × 7 × 32

= 1408 cm^{2}

**6. A rectangular sheet of paper, 44 cm× 20 cm, is rolled along its length to form a cylinder. Find the total surface area of the cylinder thus generated.**

**Solution:**

We have,

Dimensions of rectangular sheet of paper = 44cm × 20cm

When this sheet of paper is rolled along its length,

Circumference of base becomes = 44 cm

By using the formula,

Circumference of base = 2πr

So, 2πr = 44

2 × 22/7 × r = 44

r = 44×7 / 44

= 7cm

Radius = 7cm

Height = 20cm

∴ Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7 (20 + 7)

= 2 × 22/7 × 7 × 27

= 1188 cm^{2}

**7. 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:**

We have,

Ratio of radius of two cylinder, r_{1}:r_{2} = 2:3

Ratio of their heights, h_{1}:h_{2} = 5:3

r_{1}/r_{2} = 2/3

h_{1}/h_{2} = 5/3

so,

Curved surface area of cylinder1 / curved surface area of cylinder2 = 2πr_{1}h_{1} / 2πr_{2}h_{2}

= (2 × 22/7 × 2 × 5) / (2 × 22/7 × 3 × 3)

= 10/9

∴ Ratio of their curved surface area is 10:9

**8. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Prove that its height and radius are equal.**

**Solution:**

We have,

Let radius of cylinder be ‘r’

Let height of cylinder be ‘h’

Curved surface area of cylinder / total surface area of cylinder = 1/2

2πrh / 2πr (h+r) = 1/2

h/(h+r) = 1/2

2h = h+r

2h – h = r

h = r

Height = Radius

Hence proved.

**9. The curved surface area of a cylinder is 1320 cm ^{2} and its base has diameter 21 cm. Find the height of the cylinder.**

**Solution:**

We have,

Diameter of base = 21 cm

Radius of cylinder = 21/2 cm

Let height of cylinder be ‘h’ cm

Curved surface area of cylinder = 1320 cm^{2}

By using the formula,

Curved surface area of cylinder = 2πrh

So,

2πrh = 1320

2 × 22/7 × 21/2 × h = 1320

66h = 1320

h = 1320/66 = 20cm

∴ Height of cylinder is 20cm.

**10. The height of a right circular cylinder is 10.5 cm. If three times the sum of the areas of its two circular faces is twice the area of the curved surface area. Find the radius of its base.**

**Solution:**

We have,

Height of cylinder = 10.5 cm

Let radius of cylinder be ‘r’ cm

So,

Area of two bases of cylinder = 2πr^{2}

Area of curved surface of cylinder = 2πrh

Now,

3 (2πr^{2}) = 2 (2πrh)

6πr^{2} = 4πrh

πr^{2}/ πr = 4/6 h

r = 2/3 h

= 2×10.5 /3

= 7cm

∴ Radius of base of cylinder is 7cm.

**11. Find the cost of plastering the inner surface of a well at Rs 9.50 per m ^{2}, if it is 21 m deep and diameter of its top is 6 m.**

**Solution:**

We have,

Height of cylinder = 21m

Diameter of cylinder = 6m

Radius of cylinder = 6/2 = 3m

Curved surface area of cylinder = 2πrh

= 2 × 22/7 × 3 × 21

= 396 m^{2}

∴ Cost of plastering the inner surface at the rate of Rs 9.50 per m^{2} = 396 × 9.50 = Rs 3762

**12. A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.**

**Solution:**

We have,

Diameter of base of cylinder = 20 cm

Radius of cylinder = 20/2 = 10cm

Height of cylinder = 14 cm

Total surface area of cylinder = 2πrh + πr^{2}

= (2 × 22/7 × 10 × 14) + (22/7 × 10^{2})

= 880 + 2200/7

= (6160+2200)/7

= 8360/7 cm^{2}

We know that cost per 100cm^{2} = 50paise

So, cost per 1cm^{2} = Rs 0.005

∴ Cost of tin painting the area inside the vessel = 8360/7 × 0.005

= Rs 5.97

**13. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find the cost of plastering its inner curved surface at Rs. 4 per square metre.**

**Solution:**

We have,

Inner diameter of circular well = 3.5 m

Radius of well = 3.5/2 m

Height of well = 10 m

So,

Curved surface area of well = 2πrh

= 2 × 22/7 × 3.5/2 × 10

= 110 m^{2}

We know, Cost of plastering 1m^{2} = Rs 4

∴ Cost of plastering its inner curved surface of area 110m^{2} = 110 × 4 = Rs 440

**14. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions moving once over to level a playground. What is the area of the playground?**

**Solution:**

We have,

Diameter of roller = 84 cm

Radius of roller = 84/2 = 42cm

Length of roller = 120 cm

Curved surface area of roller = 2πrh

= 2 × 22/7 × 42 × 120

= 31680 cm^{2}

It takes 500 complete revolutions to level the playground.

∴ Area of playground = 500 × 31680

= 15840000 cm^{2}

= 1584 m^{2}

**15. Twenty one cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m, what will be the cost of cleaning them at the rate of Rs. 2.50 per square metre?**

**Solution:**

We have,

Number of pillars = 21

Diameter of pillar = 0.50 m

Radius of pillar = 0.50/2 = 0.25m

Height of pillar = 4 m

Rate of cleaning = Rs 2.5 per m^{2}

Curved surface area of pillar = 2πrh

= 2 × 22/7 × 0.25 × 4

= 44/7 m^{2}

Curved surface area of 21 pillars = 21 × 44/7 = 132m^{2}

∴ Cost of cleaning the pillars at the rate of R 2.50 per m^{2} = 2.50 × 132 = Rs 330

**16. The total surface area of a hollow cylinder which is open from both sides if 4620 sq. cm, area of base ring is 115.5 sq. cm. and height 7 cm. Find the thickness of the cylinder.**

**Solution:**

We have,

Total surface area of hollow cylinder = 4620 cm^{2}

Area of base ring = 115.5 cm^{2}

Height of cylinder = 7 cm

Let outer radius be ‘R’ cm , inner radius be ‘r’ cm

Area of hollow cylinder = 2π(R^{2} – r^{2}) + 2πRh + 2πrh

= 2π(R+r)(R-r) + 2πh(R+r)

= 2π(R+r) (h+R-r)

Area of base = πR^{2} – πr^{2}

= π (R^{2} – r^{2})

= π (R+r) (R-r)

Surface area / area of base = 4620/115.5

{2π(R+r) (h+R-r)} / { π (R+r) (R-r)} = 4620/115.5

2(h+R-r) / (R-r) = 4620/115.5

Let us consider (R-r) = t

2(h+t)/t = 40

2h + 2t = 40t

2h = 38t

2(7) = 38t

14 = 38t

t = 14/38

= 7/19 cm

∴ Thickness of cylinder is 7/19cm

**17. The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m ^{2}, find the circumference of its base.**

**Solution:**

We have,

Sum of base radius and height of cylinder = 37m

(r + h) = 37m

Total surface area = 1628 m^{2}

By using the formula,

Total surface area = 2πr(h+r)

So,

2πr(h+r) = 1628

2 × 22/7 × r (37) = 1628

r = 1628×7 / 2× 22×37

= 11396 / 1628

= 7m

∴ Circumference of its base = 2πr

= 2 × 22/7 × 7

= 44m

**18. Find the ratio between the total surface area of a cylinder to its curved surface area, given that its height and radius are 7.5 cm and 3.5 cm.**

**Solution:**

We have,

Radius of cylinder = 3.5 cm

Height of cylinder = 7.5 cm

By using the formulas,

Total surface area = 2πr (h+r)

Curved surface area = 2πrh

So, from the question,

Total surface area / Curved surface area = 2πr (h+r) / 2πrh

= (h+r)/h

= (3.5+7.5)/7.5

= 11/7.5

= 110/75

= 22/15

∴ Ratio between the two areas is 22:15

**19. A cylindrical vessel, without lid, has to be tin-coated on its both sides. If the radius of the base is 70 cm and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs. 3.50 per 1000 cm ^{2}.**

**Solution:**

We have,

Radius of base = 70 cm

Height of base = 1.4 m = 140 cm

Rate of tin plating = Rs 3.50 per cm^{2}

So,

Total surface area of vessel = Area of outer side of base + Area of inner and outer curved surface

= 2 (πr^{2} + 2 πrh)

= 2πr (r + 2h)

= 2 × 22/7 × 70 (70 + 2(140))

= 2 × 22/7 × 70 (350)

= 154000 cm^{2}

∴ Cost of tin coating at the rate of Rs 3.50 per 1000 cm^{2} = 3.50/1000 × 154000

= Rs 539

**All Chapter RD Sharma Solutions For Class 8 Maths**

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