## NCERT solutions for class 9 Maths

## Surface Areas and Volumes

**Chapter 13**

**Exercise 13.2**

**EX 13.2 QUESTION 1.**

**The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder.**

**Solution:**

Let the radius of base of cylinder. = r cm

height (h) = 14 cm and curved surface area of cylinder = 88 cm

^{2}

Curved surface area of a cylinder = 2πrh

⇒ 2πrh = 88

⇒ 2 x

**x r x 14 = 88**

⇒ r =

**= 1 cm**

∴ Diameter = 2 x r = (2 x 1) cm = 2 cm

**Ex 13.2 Question 2.**

**It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?**

**Solution:**

Height (h) = 1 m

Diameter of the base = 140 cm = 1.40 m

Radius (r) = m = 0.70 m

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

= 2 x x 0.70(1 + 0.70)m^{2}

= 2 x 22 x 0.10 x 1.70 m^{2}

= 2 x 22 x x m^{2}

= m^{2} = 7.48 m^{2}

Hence, the required sheet = 7.48 m^{2}

**Ex 13.2 Question 3.**

**A metal pipe is 77 cm long. The inner ft diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its**

**(i) inner curved surface area.**

**(ii) outer curved surface area.**

**(iii) total surface area.**

**Solution:**

Height (h)= 77 cm

Diameter = 4 cm I

Inner radius (r) = cm = 2 cm

Outer diameter = 4.4 cm

⇒ Outer radius (R) = cm = 2.2 cm

(i) Inner curved surface area = 2πrh

= 2 x x 2 x 77 cm^{2}

= 2 x 22 x 2 x 11 cm^{2} = 968 cm^{2}

(ii) Outer curved surface area = 2πRh

= 2 x x 2.2 x 77 cm^{2}= 2 x 22 x 2.2 x 11 cm^{2} = 1064.80cm^{2}

(iii)Inner radius of pipe r = 2cm , Outer radius R = 2.2 cm and Height h = 77 m

Area of upper ring of pipe = π(R^{2} – r^{2})

22/7 * (2.2^{2} -2^{2}) = 22/7 × (4.84 -4) = 22 × 0.12 = 2.64cm^{2}

Area of lower ring of pipe = 2.64 cm^{2}

Total surface area = [Inner curved surface area] + [Outer curved surface area] + [Area of two circular ends]=

968 + 1064.80 + 2.64 + 2.64 = 2038.08 cm²

**Ex 13.2 Question 4.**

**The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m ^{2}.**

**Solution:**

The roller is in the form of a cylinder of diameter = 84 cm

⇒ Radius of the roller(r) = cm = 42 cm = 0.42 m

Length of the roller (h) = 120 cm =1.2 m

Curved surface area of the roller = 2πrh

= 2 x x 0.42 x 1.2m

^{2}

= 2 x 22 x 0.06 x 1.2 m

^{2}= 3.168m

^{2}

Area of the playground levelled in one revolution = 3.168m^{2}

∴ Area of the playground levelled in 500 revolutions = 500 x 3.168m^{2} = 1584m^{2}

**Ex 13.2 Question 5.**

**A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m ^{2} .**

**Solution:**

Diameter of the pillar = 50 cm

∴ Radius (r) = cm = 25 cm = 0.25

and height (h) = 3.5m

Curved surface area of a pillar = 2πrh

=2× × 0.25 ×3.5 = 2× 22× 0.25 × 0.5 = 5.5 m ²

∴ Cost of painting of 1 m^{2} area = Rs. 12.50

∴ Cost of painting of 5.5 m^{2} area= ₹ ( 5.5x 12.50 )

= ₹68.75.

**Ex 13.2 Question 6.**

**Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.**

**Solution:**

Radius (r) = 0.7 m

Curved surface area of a cylinder = 2πrh

4.4 = 2 x x x h

4.4= 4.4h

h = 1m

**Ex 13.2 Question 7.**

**The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find**

**(i) its inner curved surface area.**

**(ii) the cost of plastering this curved surface at the rate of ₹40 per m ^{2}.**

**Solution:**

Radius of the well (r) = m = 1.75 m

Height of the well (h) = 10 m

(i) Inner curved surface area = 2πrh

=2 * * 1.75 * 10 = 110 m

^{2}

(ii) The cost of plastering this curved surface at the rate of per m^{2}.= ₹ 40

∴ Total cost of plastering the area 110 m^{2 }= ₹(110 x 40) = ₹4400

**Ex 13.2 Question 8.**

**In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.**

**Solution:**

Length (h) = 28 m

∴ Radius (r) = cm = 2.5m

Total surface area of a cylindrical pipe = 2πr(r + h)

= 2* * 0.025 * (0.025 + 28) = 4.4 m^{2} (approx)

Thus, the total radiating surface in the system is 4.4 m2 .

**Ex 13.2 Question 9.**

**Find**

**(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.**

**(ii) how much steel was actually used, if of the steel actually used was wasted in making the tank.**

**Solution:**

Radius (r) = = 2.1 m

Height (h) = 4.5 m

(i) Curved surface area of a cylindrical petrol tank = 2πrh

= 2 x x 2.1 x 4.5 m^{2}

= 2 x 22 x 0.3 x 4.5 m^{2} 59.4 m^{2}

(ii) Total surface area of the tank = 2πr(r + h)

= 2 x x 2.1(2.1 + 4.5)m^{2}

= 44 x 0.3 x 6.6 m^{2} = 87.13 m^{2}

Let the area of the steel used to make this cylindrical petrol storage tank = x m^{2}

∴ Area of steel that was wasted = x x m = m^{2}

Area of steel used = x – m^{2}

Thus, the required area of the steel that was actually used is 95.04 m^{2}.

**Ex 13.2 Question 10.**

**In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.**

**Solution:**

Radius of lampshade (r) = cm = 10 cm

and height = 30 cm.

Heigh of margin for folding to cover the top and bottom of the frame. H = 30 + 2.5 + 2.5 = 35 cm

∴ Total height of the cylinder, (h)

= 30 cm + 2.5 cm + 2.5 cm = 35 cm

Now, curved surface area = 2πrH

= 2 x x 10 x 35 cm^{2}

= 2200 cm^{2}

**Ex 13.2 Question 11.**

**The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?**

**Solution:**

Radius of each penholder (r) = 3 cm

Height of a penholder (h) = 10.5 cm

The penholder open from the top,therefore the area of cardboard for 1 penholder = 2πrh + πr^{2}

= 2 * π * 3 * 10.5 + π * 3 = 63π +9π = 72π cm^{2}

∴ Surface area of 35 penholders= 35 * 72π cm^{2}

=35 * 72 * = 5 * 72 * 22 = 7920 cm^{2}