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# A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of $ Rs. 4989.60 $. If the cost of white-washing is $ Rs. 20 $ per square metre, find the

**(i)** inside surface area of the dome,

**(ii)** volume of the air inside the dome.

**Given:**

A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of \( Rs. 4989.60 \).

The cost of white-washing is \( Rs. 20 \) per square metre.

**To do:**

We have to find the

(i) inside surface area of the dome,(ii) volume of the air inside the dome.

**Solution:**

Cost of white-washing the dome $=Rs.\ 4989.60$

Rate of white-washing \( =Rs. 20 \) per square metre.

(i) Cost of white washing the dome $=\text { Surface area of the dome } \times \text { Cost of white washing per square metre }$

This implies,

Surface area of the dome $=\frac{\text { Cost of white washing the dome }}{\text { Cost of white washing per square metre }}$

$=\frac{498.96}{2}$

$=249.48 \mathrm{sq} \mathrm{m}$

**Inside surface area of the dome is $249.48 \mathrm{sq} \mathrm{m}$.**

(ii) Let $r$ be the radius of the dome.

Surface area of the dome $=249.48 \mathrm{sq} \mathrm{m}$

This implies,

$2 \pi r^{2}=249.48$

$2 \times \frac{22}{7} \times r^{2}=249.48$

$r^{2}=\frac{249.48 \times 7}{44}$

$r^{2}=36.69$

$r=\sqrt{36.69}$

$r=6.3 \mathrm{~m}$

Therefore,

Volume of the air inside the dome $=$ Volume of the dome

$=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times(6.3)^{3}$

$=523.908\ m^3$

**The volume of the air inside the dome is $523.908\ m^3$.**

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