9th Maths 13.2
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.1
NCERT Class 9th solution of Exercise 13.3
NCERT Class 9th solution of Exercise 13.4
NCERT Class 9th solution of Exercise 13.5
NCERT Class 9th solution of Exercise 13.6
NCERT Class 9th solution of Exercise 13.7
NCERT Class 9th solution of Exercise 13.8
Exercise 13.2
Assume π=227, unless stated otherwise.
Q1. The curved surface area of a right circular cylinder of height 14cm is 88cm2. Find the diameter of the base of the cylinder.
Sol. :
Given:
Height of Cylinder is h=14cm,
Area=88cm2
To Find:
Diameter of the base of Cylinder.
Solve:
Let Radius is r
Curved Surface Area=2πrh
88=2×227×⋊
r=(88times7)/(2times22times14)
r=1
Diameter of the base=2r=2times1=2cm
Answer:
Diameter of the base 2cm.
Q2. It is required to make a closed cylindrical tank of height 1m and base diameter 140cm from a metal sheet. How many square meters of the sheet are required for the same?
Sol. :
Given:
Height of the Cylinder is h=1m,
Diameter of base isD=140cm=1.40m, r=(1.40)/2=0.70m
To Find:
Area of the sheet.
Solve:
Let Radius r, and height h.
Area of closed Cylinder =Total Surface Area
=2pir(r+h)
=2times22/7times0.7times(1+0.7)
=2times22times0.1times1.70
=7.48m^2
Answer:
Area of the sheet 7.48m^2.
Q3. A metal pipe is 77cm long. The inner diameter of a cross-section is 4cm the outer diameter being 4.4cm (see figure). Find its
i) inner curved surface area,
ii) outer curved surface area,
iii) total surface area.
Sol. :
Given:
External radius R=(4.4)/2=2.2cm
Internal radius r=(4)/2=2cm
Length of the pipe h=77cm
To Find:
i) inner curved surface area,
ii) outer curved surface area,
iii) total surface area.
Solve:
i)
Inner curved Surface Area=2pirh
=2times(22)/7times2times77
=968cm^2
Answer:
Inner curved Surface Area is 968cm^2.
ii)
Outer Curved Surface Area=2piRh
=2times(22)/7times2.2times77
=1064.8cm^2
Answer:
Outer Curved Surface Area is 1064.8cm^2.
iii)
Total Surface Area of a pipe
=Inner Curved Surface Area + Outer Curved Surface Area
+ Areas of two bases
=2pirh + 2piRh+2pi(R^2-r^2)
= 968+1064.8+2times(22)/7(4.48-4)
=2032.8+(44)/7times0.84
=2032.8+5.28
=2038.08cm^2
Answer:
Total Surface Area of a pipe is 2038.08cm^2.
Q4. The diameter of a roller is 84cm and its length is 120cm. It takes 500 complete revolutions to move once to level a playground. Find the area of the playground in m^2.
Sol. :
Given:
Radius of roller r=84/2=42cm
Length of roller h=120cm=1.2m
To Find:
Area of the playground.
Solve:
Distance covered by roller in One revolution
= Curved Surface Area of roller
=2pirh
=2times(22)/7times0.42times1.2
=3.168m^2
Area of playground
= Distance covered by roller in 500 revolution
=500times3.168m^2
=1584m^2
Answer:
Area of playground 1584m^2.
Q5. A cylindrical pillar is 50cm in diameter and 3.5m in height. Find the cost of painting the curved surface of the pillar at the rate of ₹12.50 per m^2.
Sol. :
Given:
Radius of Cylinder is r=(50)/2=25cm=0.25m
Height of Cylinder is h=3.5m
Rate of paint ₹12.50 per m^2
To Find:
Cost of paint.
Solve:
Curved Surface Area of Cylinder
=2pirh
=2times(22)/7times0.25times3.5
=5.5m^2
Cost of paint = AreatimesRate
=5.5times12.50
=68.75
Answer:
Cost of paint ₹68.75.
Q6. Curved surface area of a right circular cylinder is 4.4m^2. If the radius of the base of the cylinder is 0.7m, find its height.
Sol. :
Given:
Curved Surface Area of the Cylinder is 4.4m^2
Radius of the base is 0.7m
To Find:
Height of the Cylinder.
Solve:
Curved Surface Area =2pirh
4.4=2times(22)/7times0.7timesh
h=(44times7)/(2times22times0.7)=1m
Answer:
The height of the Cylinder is 1m.
Q7. The inner diameter of a circular well is 3.5m. It is 10m deep. Find
i) its inner curved surface area,
ii) the cost of plastering this curved surface at the rate of ₹40 per m^2.
Sol. :
Given:
Inner radius is r=(3.5)/2m
Height of well h=10m
Rate ₹40 per m^2.
To Find:
i) its inner curved surface area,
ii) the cost of plastering this curved surface at the rate of ₹40 per m^2.
Solve:
i)
Inner Curved Surface Area =2pirh
=2times(22)/7times(3.5)/2times10=110m^2
Answer:
inner Curved Surface Area is 110m^2.
ii)
Cost of Plastering = AreatimesRate
=110times40
=₹4400
Answer:
Cost of plastering ₹4400.
Q8. In a hot water heating system, there is a cylindrical pipe of length 28m and diameter 5cm. Find the total radiating surface in the system.
Sol. :
Given:
Length of Cylinder is h=28m
Radius of Cylinder is r=5/2=2.5cm=0.025m
To Find:
Total radiating Surface Area of the system
Solve:
Total radiating Surface Area of the system
= Curved Surface area of the Cylinder
=2pirh
=2times(22)/7times0.025times28
=4.4m^2
Answer:
Total radiating Surface Area of the system is 4.4m^2.
Q9. Find
i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2m in diameter and 4.5m high.
ii) how much steel was actually used if 1/12 of the steel actually used was wasted in making the tank.
Sol. :
Given:
Height of the closed Cylinder is 4.5m
Radius of the closed Cylinder is (4.2)/2m=2.1m
To Find:
i) Curved Surface Area of the closed Cylinder.
ii) How much steel used.
Solve:
i)
Curved Surface Area of Cylinder
=2pirh
=2times(22)/7times2.1times4.5
=59.4m^2
ii)
Let Actual Area of Steel used is x
(1-1/(12))x=2pirh+2pir^2
(11)/(12)x=59.4+2times(22)/7times2.1times2.1
(11)/(12)x=59.4+27.72
(11)/(12)x=87.12
x=(87.12times12)/11
x=95.04m^2
Answer:
The Actual Area of the Steel used 95.04m^2.
Q10. 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 20cm and height of 30cm. A margin of 2.5cm 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.
Sol. :
Given:
Radius of frame base is r=(20)/2=10cm
Top and Bottom margin =2.5+2.5=5cm
Height of frame is h=30+5=35cm
To find:
Area of Cloth.
Solve:
Area of Cloth = Curved Surface Area of Cylinder
=2pirh
=2times(22)/7times10times35
=2200cm^2
Answer:
Area of Cloth 2200cm^2.
Q11. 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 3cm and height 10.5cm. 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?
Sol. :
Given:
Radius of penholder is r=3cm
Height of penholder is h=10.5cm
Number of Competitors is 35
To Find:
Area of Cardboard
Solve:
Area of Cardboard
= Curved Surface Area of penholder + Area of base
=2pirh+pir^2
=2times(22)/7times3times10.5+(22)/7times9
=198+28.28
=226.28cm^2 [approx]
Cardboard for 35 competitors
=35times226.28
=7920cm^2
Answer:
The area of the Cardboard for the 35 competitors were 7920cm^2.
Comments