Gujarat Board GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes 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 the base of the cylinder be r cm.

h = 14 cm

Curved surface area = 88 cm^{2}

2πrh = 88

= 2 x \(\frac {22}{7}\) x r x 14 = 88

r = \(\frac{88 \times 7}{2 \times 22 \times 14}\)

r = 1 ⇒ 2r = 2

Hence the diameter of the base of the cylinder is 2 cm.

Question 2.

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

Solution:

h = 1 m = 100 cm

2r = 140 cm

r = \(\frac {140}{2}\)cm = 70 cm

∴ Total surface area of the closed cylindrical tank

= 2πr(h + r)

= 2 x \(\frac {22}{7}\) x 70 (100 + 70)

= 74800 cm^{2} = \(\frac {74800}{100 x 100}\) m^{2} = 7.48 m^{2}

Hence, 7.48 square metres of the sheet are required.

Question 3.

A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. Find its

1. inner curved surface area,

2. outer curved surface area,

3. the total surface area.

Solution:

h = 77 cm

2r = 4 cm

⇒ r = 2 cm

2R = 4.4 cm

⇒ R = 2.2 cm

1. Inner curved surface area = 2πrh

= 2 x \(\frac {22}{7}\) x 2 x 77 = 968 cm^{2}

2. Outer curved surface area = 2πRh

= 2 x \(\frac {22}{7}\) x 2.2 x 77 = 1064.8 cm^{2}

3. Total surface area

= 2πRh + 2πrh + 2π(R^{2} – r^{2})

= 1064.8 + 968 + 2 x \(\frac {22}{7}\) [(2.2)^{2} – (2)^{2} ]

= 1064.8 + 968 + 2 x \(\frac {22}{7}\) x (4.84 – 4)

= 1064.8 + 968 + 2 x \(\frac {22}{7}\) x 0.84 7

= 1064.8 + 968 + 5.28 = 2038.08 cm^{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:

2r = 84 cm

r = 42 cm

h = 120 cm

∴ Area of the playground levelled in taking 1 complete revolution

= 2πrh = 2 x \(\frac {22}{7}\) x 42 x 120

= 31680 cm^{2}

∴ Area of the playground

= 31680 x 500 = 15840000 cm^{2}

= \(\frac{15840000}{100 \times 100}\) m^{2} = 1584 m^{2}

Hence, the area of the playground is 1584 m^{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:

2r = 50 cm

∴ r = 25 cm = 0.25 m

h = 3.5 m

∴ Curved surface area of the pillar

= 2πrh = 2 x \(\frac {22}{7}\) x 0.25 x 3.5

= 5.5 m^{2}

∴ Cost of painting the curved surface area of the pillar at the rate of ₹ 12.50 per m^{2}

= ₹ (5.5 x 12.50) = ₹ 68.75

Question 6.

The curved surface area of a right circular cylinder is 4.4 m^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:

Let the height of the right circular cylinder be h m.

r = 0.7 m

Curved surface area = 4.4 m^{2}

2πrh = 4.4

2 x \(\frac {22}{7}\) x 0.7 x h = 4.4

4.4h = 4.4 = h = 1m

Hence, the height of the right circular cylinder is 1 m.

Question 7.

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

1 its inner curved surface area,

2. the cost of plastering this curved surface at the rate of 40 per m^{2}.

Solution:

1. 2r = 3.5 m

r = \(\frac {3.5}{2}\) m

= r = l.75 m

h = 10m

∴ Inner curved surface area of the circular wall = 2πrh

= 2 x \(\frac {22}{7}\) x 1.75 x 10 = 110 m^{2}.

2. Cost of plastering the curved surface at the rate of ₹ 40 per m^{2}

= ₹ (110 x 40) = ₹ 4400

Question 8.

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

Solution:

h = 28m

2r = 5 cm

r = \(\frac {5}{2}\)cm = x \(\frac {5}{2 x 100}\) m

∴ Total radiating surface in the system

= 2πrh = 2 x \(\frac {22}{7}\) x \(\frac {1}{40}\) x 28

= 4.4 m^{2}

Question 9.

Find:

1. 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.

2. how much steel was actually used, if of the steel actually used was wasted in making the tank.

Solution:

1. 2r = 4.2 m

∴ r = \(\frac {4.2}{2}\) m = 2.1 m

h = 4.5m

∴ Lateral or curved surface area = 2πrh

= 2 x \(\frac {22}{7}\) x 2.1 x 4.5 = 59.4 m^{2}

2. Total surface area = 2πr(h + r)

= 2 x \(\frac {22}{7}\) x 2.1 x (4.5 + 2.1)

= 2 x \(\frac {22}{7}\) x 2.1 x 6.6 = 87.12 m^{2}

Let the actual area of steel used be x m^{2}.

Since \(\frac {1}{12}\) of the actual steel used was wasted, the area of the steel has gone into the tank = \(\frac {11}{12}\) of x.

∴ \(\frac {11}{12}\) x = 87.12

∴ x = \(\frac{87.12 \times 12}{11}\) = 95.04m^{2}

∴ Steel actually used = 95.04 m^{2}.

Question 10.

In the figure given below, 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:

2r = 20 cm r = 10cm

h = 30cm

∴ Cloth required = 2,tr (h + 2.5 + 2.5)

= 2πr (h + 5)

= 2 x \(\frac {22}{7}\) x 10 x (30+5)

= 2200 cm^{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?

r = 3 cm

h = 10.5 cm

∴ Cardboard required for 1 competitor

= 2πrh + πr^{2}

= 2 x \(\frac {22}{7}\) x 3 x 10.5 + \(\frac {22}{7}\) (3)^{2}

= 198 + \(\frac {198}{7}\) = 198(1+ \(\frac {1}{7}\))

= \(\frac {198 x 8}{7}\) cm^{2}

∴ Cardboard required for 35 competitors

= \(\frac {198 x 8}{7}\) x 35cm^{2} = 7920 cm^{2}

Hence, 7920 cm^{2} of cardboard was required to be bought for the competition.