9th Maths 13.4
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 13.5
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 13.4
Assume `pi=(22)/7`, unless stated otherwise.
Q1. Find the surface area of a sphere of radius:
i) `10.5cm` ii) `5.6cm` iii) `14cm`
Sol. :
i) Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times(10.5)^2`
`=1386cm^2`
Answer:
Surface Area of a Sphere `1386cm^2`.
ii) Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times(5.6)^2`
`=394.24cm^2`
Answer:
Surface Area of a Sphere `394.24cm^2.`
iii) Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times(14)^2`
`=2464cm^2`
Answer:
Surface Area of a Sphere `2464cm^2`.
Q2. Find the surface area of a sphere of diameter:
i) `14cm` ii) `21cm` iii) `3.5m`
Sol. ;
i)
Given:
Radius of Sphere is `r=(14)/2=7`
Solve:
Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times(7)^2`
`=616cm^2`
Answer:
Surface Area of a Sphere `616cm^2`.
ii)
Given:
Radius of Sphere is `r=(21)/2`
Solve:
Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times((21)/2)^2`
`=1386cm^2`
Answer:
Surface Area of a Sphere `1386cm^2`.
iii)
Given:
Radius of Sphere is `r=(3.5)/2`
Solve:
Surface Area of a Sphere `=4pir^2`
`=4times(22)/7times((3.5)/7)^2`
`=38.5cm^2`
Answer:
Surface Area of a Sphere `38.5cm^2`.
Q3. Find the total surface area of a hemisphere of radius `10cm`. (Use `pi=3.14`)
Sol. :
Given:
Radius of a hemisphere is `r=10cm`.
To Find:
Total Surface Area of a Hemisphere.
Solve:
Total Surface Area of a Hemisphere `=3pir^2`
`=3times3.14times(10)^2`
`=942cm^2`
Answer:
Total Surface Area of a Hemisphere `942cm^2`.
Q4. The radius of a spherical balloon increases from `7cm` to `14cm` as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Sol. :
Given:
Radius of Spherical balloon `r_1=7cm, r_2=14cm`
To Find:
Ratio of Surface Areas.
Solve:
Ratio of Surface Areas `=(4pi(r_1)^2)/(4pi(r_2)^2)`
`=(7)^2/(14)^2=(7/14)^2=(1/2)^2=1/4`
Answer:
Ratio of Surface Areas of the balloon in the two cases is `1:4`.
Q5. A hemispherical bowl made of brass has inner diameter `10.5cm`. Find the cost of tin-plating it on the inside at the rate of `₹16`per`100m^2`.
Sol. :
Given:
Radius of Hemisphere `r=(10.5)/2=5.25cm`
Rate `₹16` per `100m^2`
To Find:
Cost of tin-plating.
Solve:
Curved Surface Area of Hemisphere `=2pir^2`
`=2times(22)/7times(5.25)^2`
`=173.25cm^2`
Cost of tin-plating `=AreatimesRate`
`=₹173.25times(16)/(100)`
`=₹27.72`
Answer:
Cost of tin-plating `₹27.72`.
Q6. Find the radius of a sphere whose surface area is `154cm^2`.
Sol. :
Given:
Surface Area of Sphere `154cm^2`.
To Find:
Radius
Solve:
Surface Area of Sphere `=4pir^2`
`154=4times(22)/7timesr^2`
`r^2=(154times7)/22`
`r^2=7times7`
`r=7cm`
Answer:
Radius of Sphere `7cm`.
Q7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Sol. :
Given:
Radius of earth is `r_e=D/2`
Radius of moon is `r_m=D/(2times4)=D/8`
To Find:
Ratio of Surface Area.
Solve:
Ratio of Surface Area `=(4pi(r_m)^2)/(4pi(r_e)^2`
`=(r_m^2)/(r_e^2)`
`=(D/8)^2/(D/2)^2`
`=(1/64)/(1/4)=(1/64)times(4/1)=4/64=1/16`
Answer:
Ratio of their Surface Areas `1:16`.
Q8. A hemispherical bowl is made of steel, `0.25cm` thick. The inner radius of the bowl is `5cm`. Find the outer curved surface of the bowl.
Sol. :
Given:
Inner Radius `r=5cm.`
Thickness of steel `=0.25cm`
Outer Radius `R=5+0.25=5.25cm`
To Find:
Outer Curved Surface Area of Bowl.
Solve:
Curved Surface Area of Hemisphere = Outer Curved Surface Area of Bowl
`=2piR^2`
`=2times(22)/7times(5.25)^2`
`=173.25cm^2`
Answer:
Outer Curved Surface Area of Bowl `173.25 cm^2`.
Q9. A right circular cylinder just encloses a sphere of radius `r`(see figure). Find
i) surface area of the sphere,
ii) curved surface area of the cylinder,
iii) ratio of the areas obtained in i) and ii).
Sol. :
i) Surface Area of the Sphere `=4pir^2`
ii)
`Radius=r,` and `Height h=2r`
Curved Surface Area of the Cylinder `=2pirh=2pi r times 2r`
`=4pir^2`
Answer:
Curved Surface Area of the Cylinder `=4pir^2`.
iii)
Ratio of the Areas `=(4pir^2)/(4pir^2)=1/1`
Answer:
Ratio of the Areas `1:1`.
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