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 `pi=22/7`, unless stated otherwise.
Q1. The curved surface area of a right circular cylinder of height `14cm` is `88cm^2`. Find the diameter of the base of the cylinder.
Sol. :
Given:
Height of Cylinder is `h=14cm,`
` Area=88cm^2`
To Find:
Diameter of the base of Cylinder.
Solve:
Let Radius is `r`
Curved Surface Area`=2pirh`
`88=2times22/7timesrtimes14`
`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 is`D=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 = Area`times`Rate
`=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 `=` Area`times`Rate
`=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`.
 

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