Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.2 Textbook Exercise Questions and Answers.
Haryana Board 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.2
Assume π = \(\frac{22}{7}\), unless stated otherwise.
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:
We have, height of the cylinder (h) = 14 cm
Let radius of the cylinder be r cm and curved surface area of the cylinder = 88 cm2
⇒ 2πrh = 88
⇒ 2 × \(\frac{22}{7}\) × r × 14 = 88
⇒ r = \(\frac{88 \times 7}{2 \times 22 \times 14}\)
⇒ r = 1 cm
Diameter = 2 × 5 = 2 × 1 = 2 cm
Hence,diameter of the base of the cylinder = 2 cm.
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:
We have,
Base diameter of the cylinder (d) = 140 cm
∴ Radius of the cylinder (r) = \(\frac{140}{2}\) = 70 cm
and height of the cylinder (h) = 1 m = 100 cm
Required metal sheet to make a closed cylindrical tank = Total surface area of the cylinder
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 70(100 + 70)
= 440(170) = 74800 cm2
= \(\frac{74800}{10000}\) m2 = 7.48 m2
Hence, area of the required sheet = 7.48 m2.
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 (see Fig. 13.15). Find its : (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area.
Solution:
We have, Length of cylindrical pipe (h) = 77 cm
Inner diameter of the cross section = 4 cm.
∴ Inner radius of the cross section (r) = \(\frac{4}{2}\)
= 2 cm
and outer diameter of the cross section = 4.4 cm
∴ outer radius of the cross section
(R) = \(\frac{4.4}{2}\) = 2.2 cm
(i) Inner curved surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77
= 968 cm2
(ii) Outer curved surface area = 2πRh
= 2 × \(\frac{22}{7}\) × 2.2 × 77
= 44 × 2.2 × 11
= 1064.8 cm2
(iii) Surface area of the two bases = 2 × π
(R2 – r2) = 2 × \(\frac{22}{7}\) [(2.2)2 – (2)2]
= \(\frac{44}{7}\)(2.2 + 2) (2.2 – 2)
= \(\frac{44}{7}\) × 4.2 × 0.2
= 44 × 0.6 × 0.2
= 5.28 cm2
∴ Total surface area= Inner curved surface area + outer curved surface area + surface area of the two bases
= 968 + 1064.8 +5.28
= 2038.08 cm2
Hence, inner C.S.A., outer C.S.A. and T.S.A. of the metal pipe are 968 cm2, 1064.8 cm2 and 2038.08 cm2 respectively.
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:
We have,
length of roller (h)= 120 cm
and diameter of roller = 84 cm
∴ Radius of the roller (r) = \(\frac{84}{2}\) = 42 cm
Area covered by roller in 1 revolution = curved surface area of roller
= 2πrh
= 2 × \(\frac{22}{7}\) × 42 × 120
= 22 × 12 × 120
= 31680 cm2
Area covered by roller in 500 revolutions = 31680 × 500
= 15840000 cm2
= \(\frac{15840000}{10000}\) m2
= 1584 m2
Hence,area of the playground = 1584 m2.
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 Rs. 12.50 per m2.
Solution:
We have, Height of the cylindrical pillar (h) = 3.5 m
= 350 cm
and diameter of the cylindrical pillar = 50cm
∴ Radius of the cylindrical pillar (r) = \(\frac{50}{2}\)
= 25 cm
Curved surface area of the cylindrical pillar
= 2πrh
= 2 × \(\frac{22}{7}\) × 25 × 350
= 44 × 25 × 50
= 55000 cm2
= \(\frac{55000}{10000}\) m2 = 5.5 m2
Rate of the painting = Rs. 12.50 per m2
Cost of painting the curved surface area of the pillar
= 5.5 × 12.50 = Rs. 68.75
Hence, cost of the painting the C.S.A. of the pillar = Rs. 68.75.
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.
Solution:
We have,
Radius of the circular cylinder (r) = 0.7 m
Let height of the circular cylinder be h m andcurved surface area of a right circular cylinder= 4.4 m2
⇒ 2πrh = 4.4
⇒ 2 × \(\frac{22}{7}\) × 0.7 × h = 4.4
⇒ \(\frac{44}{10}\) × h = 4.4
⇒ h = \(\frac{4.4 \times 10}{44}\)
⇒ h = 1m
Hence, height of the right circular cylinder = 1m.
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 Rs. 40 per m2.
Solution:
We have,
Depth of the circular well (h) = 10 m
and inner diameter of the circular well (d) = 3.5 m
∴ Inner radius of the circular well (r) = \(\frac{3.5}{2}\) m
Inner curved surface area of the wall = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{3.5}{2}\) × 10
= 110 m2
Rate of plastering the curved surface area of the well= Rs. 40 per m2
∴ Cost of plastering the inner curved surface of the well= 110 × 40 = Rs. 4400
Hence, (i)Inner curved surface area of the well = 110 m2
(ii) cost of plastering the inner curved surface = Rs. 4400.
Question 8.
In a hot water het ing system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
We have,
Length of cylindrical pipe (h) = 28 m
and diameter of cylindrical pipe = 5 cm
∴ Radius of cylindrical pipe (r) = \(\frac{5}{2}\) cm
= \(\frac{5}{200}\) m
The total radiating surface in the system = Curved surface area of the pipe
= 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{5}{200}\) × 28
= \(\frac{88 \times 5}{100}\) = 4.4 m2
Hence, total radiating surface in the system = 4.4 m2
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 \(\frac{1}{12}\) of the steel actually used was wasted in making the tank.
Solution:
We have, Height of cylindrical petrol tank (h) = 4.5 m
and diameter of cylindrical petrol tank = 4.2 m
∴ Radius of the cylindrical petrol tank
(r) = \(\frac{4.2}{2}\) = 2.1 m
Curved surface area of the cylindrical petrol tank = 2πrh
= 2 × \(\frac{22}{7}\) × 2.1 × 4.5
= 9 × 6.6 = 59.4 m2
Steel used to make the tank = Total surface area of the tank
= 2πr(h + r)
= 2 × \(\frac{22}{7}\) × 2.1(4.5 + 2.1)
= 13.2 × 6.6
= 87.12 m2 …(i).
Let the area of the actual steel used to make the tank be x m2.
Area of steel which wasted to make the tank
= \(\frac{1}{2}\) of x = \(\frac{x}{2}\) m2
Steel used to make the tank = x – \(\frac{x}{12}=\frac{11 x}{12}\) ….(ii)
From (i) and (ii), we get
\(\frac{11 x}{12}\) = 87.12
⇒ x = \(\frac{87\cdot12 \times 12}{11}\)
⇒ x = 95.04 m2
Hence, (i) curved surface area of the petrol tank = 59.4 m2
(ii) area of the actual steel used = 95.04 m2.
Question 10.
In Fig. 13.16, 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:
Diameter of the base of a cylindrical lampshade = 20 cm
∴ Radius of the base of a cylindrical lampshade (r) = \(\frac{20}{2}\) = 10 cm
and height of the cylindrical lampshade = 30 cm
Margin for folding over the top and bottom of the cylindrical lampshade = 2.5 cm
∴ Total height of cloth for covering the cylindrical lampshade = 30 + 2 × 2.5 = 35 cm
Area of cloth required for covering the lampshade = Curved surface area of the cylindrical lampshade
= 2πrh
= 2 × \(\frac{22}{7}\) × 10 × 35
= 220 × 10 = 2200 cm2
Hence, area of cloth required for covering the lampshade = 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?
Solution:
We have,
Radius of base of cylindrical penholder (r) = 3 cm
Height of the cylindrical pen holder (h) = 10.5 cm
Required area of cardboard for one pen holder = Total surface area
= 2πrh + πr2 [∵ penholder is open at top]
= 2 × \(\frac{22}{7}\) × 3 × 10.5 + \(\frac{22}{7}\) × 3 × 3
= \(\frac{66}{7}\)[21 +3]
= \(\frac{1584}{7}\) cm2
Required area of cardboard for 35 penholders = \(\frac{1584}{7}\) × 35
= 7920 cm2
Hence, required area of cardboard for 35 penholders = 7920 cm2.