9th Maths 13.7
Chapter 13
Surface Areas and Volumes
NCERT Class 9th solution of Exercise 13.1
NCERT Class 9th solution of Exercise 13.2
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 13.8
Exercise 13.7
Assume `pi=(22)/7`, unless stated otherwise.
Q1. Find the volume of the right circular cone with
i) radius `6cm`, height `7cm` ii) radius `3.5cm`, heirht `12cm`
Sol.
i)
Volume of the right circular cone `=1/3pir^2h`
`=1/3times(22)/7times(6)^2times7`
`=264 cm^3`
Answer:
Volume of the right circular cone `264cm^2`.
ii)
Volume of the right circular cone `=1/3pir^2h`
`=1/3times(22)/7times(3.5)^2times12`
`=154 cm^3`
Answer:
Volume of the right circular cone `154 cm^3`.
Q2. Find the capacity in litres of a conical vessel with
i) radius `7cm` slant height `25cm` ii) height `12cm,` slant height `13cm`.
Sol. :
i)
Height of the cone `h=sqrt(l^2-r^2)`
`=sqrt((25)^2-(7)^2)`
`=sqrt(625-49)`
`=sqrt576`
`=24`
Volume of the conical vessel `=1/3pir^2h`
`=1/3times(22)/7times(7)^2times24`
`=1232cm^3`
`=(1232/1000)l`
`=1.232 l`
Answer:
The capacity of the conical vessel `1.232 l`.
ii)
Radius of cone `r=sqrt(l^2-h^2)`
`r=sqrt((13)^2-(12)^2`
`r=sqrt(169-144)`
`r=sqrt(25)`
`r=5cm`
Volume of the conical vessel `=1/3pir^2h`
`=1/3times(22)/7times(5)^2times12`
`=(2200)/7cm^3`
Capacity of the conical vessel in litres
`=(2200)/7times1/(1000)=11/35 l`
Answer:
Capacity of the conical vessel `11/35 l`.
Q3. The height of a cone is `15cm`. If its volume is `1570cm^3`, find the radius of the base.
(Use `pi=3.14`)
Sol. :
Given:
Height of a cone `h=15cm`
Volume of cone `=1570cm^3`
To Find:
The radius of the base.
solve:
Volume of cone `=1/3pir^2h`
`1570 = 1/3times3.14r^2times15`
`r^2=(1570times3)/(3.14times15)`
`r^2=100`
`r=10cm`
Answer:
The radius of the base `10cm`.
Q4. If the volume of a right circular cone of height `9cm` is `48picm^3`, find the diameter of its base.
Sol. :
Given:
Height of cone `h=9cm`
Volume of cone `=48picm^3`
To Find:
The diameter of base.
Solve:
Volume of the cone `=1/3pir^2h`
`48pi=1/3pir^2times9`
`r^2=(48pi times3)/(9pi)`
`r^2=16`
`r=4cm`
Daimeter `=2times4=8cm`
Answer:
The diameter of the base of the cone `8cm`.
Q5. A conical pit of top diameter `3.5m` is `12m` deep. What is its capacity in kilolitres?
Sol. :
Given:
Radius of conical pit `r = 3.5/2 m`
Height of conical pit `h = 12 m`
To Find:
Capacity of conical pit in kilolitres.
Solve:
Volume of conical pit `=1/3pir^2h`
`=1/3times(22)/7times(3.5/2)^2times12`
`=38.5 m^3`
Capacity of conical pit `=38.5` kilolitres.
Answer:
Capacity of conical pit `38.5` kilolitres.
Q6. The volume of a right circular cone is `9856cm^3`. If the diameter of the base is `28cm,`
find
i) height of the cone ii) slant height of the cone
iii) curved surface area of the cone.
Sol. :
Given:
The volume of cone `9856cm^3`
The radius of the base `r=28/2=14cm`
To find:
i) height of the cone
ii) slant height of the cone
iii) curved surface area of the cone.
Solve:
i) Volume of cone `=1/3pir^2h`
`9856 = 1/3times(22)/7times(14)^2timesh`
`h=(9856times3times7)/(22times14times14)`
`h=48cm`
Answer:
height of the cone `48cm`
ii) Slant height of the cone `l = sqrt(h^2+r^2)`
`l = sqrt((48)^2+(14)^2)`
`l = sqrt(2304 + 196)`
`l = sqrt2500`
`l = 50cm`
Answer:
Slant height of the cone `50cm`.
iii) curved surface area of cone `=pirl`
`=(22)/7times14times50`
`=2200 cm^2`
Answer:
Curved surface area of cone `2200 cm^2`.
Q7. A right triangle `ABC` with sides `5cm, 12cm,` and `13cm` is revolved about the side `12cm`. Find the volume of the solid so obtained.
Sol. :
Given:
Radius `r=5cm`
Height `h=12cm`
To Find:
Volume of the solid.
Solve:
Volume of cone `=1/3pir^2h`
`=1/3timespitimes(5)^2times12`
`=100picm^3`
Answer:
Volume of solid `100picm^3`.
Q8. If the triangle `ABC` in the Question `7` above is revolved about the side `5cm`, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Question `7` and `8`.
Sol. :
Given:
Radius `r=12cm`
Height `h=5`
To Find:
Volume of the solid.
Solve:
Volume of cone `=1/3pir^2h`
`=1/3times(22)/7times(12)^2times5`
`=240picm^3`.
Ratio of the volume of Question `7` and `8`
`=100pi : 240pi`
`=5 : 12`
Answer:
Volume of cone `240pi,` and Ratio `5:12.`
Q9. A heap of wheat is in the form of a cone whose diameter is `10.5m` and height is `3m`. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Sol. :
Given:
Radius of cone `r=10.5/2=5.25m`
Height of cone `h=3m`
To Find:
The area of the canvas.
Solve:
Volume of cone `=1/3pir^2h`
`=1/3times(22)/7times(5.25)^2times3`
`=86.625 m^3`
Slant height `l = h^2+r^2`
`l^2 = (3)^2+(5.25)^2`
`l^2 = 9+27.5625`
`l = sqrt36.5625 = 6.0467` (approx)
The area of the canvas = Curved surface area of cone
`= pirl`
`= (22)/7times5.25times6.0467`
`=99.77 m^2` (approx)
Answer:
The area of the canvas is `99.77m^2`.
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