GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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 cm2. 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 cm2
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.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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 cm2 = \(\frac {74800}{100 x 100}\) m2 = 7.48 m2
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.
GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2
Solution:
h = 77 cm
2r = 4 cm
⇒ r = 2 cm
2R = 4.4 cm
⇒ R = 2.2 cm

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

1. Inner curved surface area = 2πrh
= 2 x \(\frac {22}{7}\) x 2 x 77 = 968 cm2

2. Outer curved surface area = 2πRh
= 2 x \(\frac {22}{7}\) x 2.2 x 77 = 1064.8 cm2

3. Total surface area
= 2πRh + 2πrh + 2π(R2 – r2)
= 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 cm2

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 m2.
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 cm2
∴ Area of the playground
= 31680 x 500 = 15840000 cm2
= \(\frac{15840000}{100 \times 100}\) m2 = 1584 m2
Hence, the area of the playground is 1584 m2.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes 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 m2.
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 m2
∴ Cost of painting the curved surface area of the pillar at the rate of ₹ 12.50 per m2
= ₹ (5.5 x 12.50) = ₹ 68.75

Question 6.
The 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:
Let the height of the right circular cylinder be h m.
r = 0.7 m
Curved surface area = 4.4 m2
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.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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 m2.
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 m2.

2. Cost of plastering the curved surface at the rate of ₹ 40 per m2
= ₹ (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 m2

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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 m2

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 m2
Let the actual area of steel used be x m2.
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 = \(\frac{87.12 \times 12}{11}\) = 95.04m2
∴ Steel actually used = 95.04 m2.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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.
GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2
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 cm2

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 + πr2
= 2 x \(\frac {22}{7}\) x 3 x 10.5 + \(\frac {22}{7}\) (3)2
= 198 + \(\frac {198}{7}\) = 198(1+ \(\frac {1}{7}\))
= \(\frac {198 x 8}{7}\) cm2
∴ Cardboard required for 35 competitors
= \(\frac {198 x 8}{7}\) x 35cm2 = 7920 cm2
Hence, 7920 cm2 of cardboard was required to be bought for the competition.

GSEB Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

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