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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 ₹4989.60. If the cost of white-washing is ₹20 per square meter, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome

**Solution:**

Surface area of a hemisphere = 2πr^{2}

Volume of a hemisphere = 2/3πr^{3}

(i) Inside surface area of the dome = Total cost for whitewashing the dome inside / Rate of whitewashing

⇒ 4989.60/20 = 249.48 m^{2}

(ii) Let 'r' be the radius of a hemispherical dome.

Inner surface area of the hemispherical dome = 2πr^{2}

2πr^{2 }= 249.48 m^{2}

⇒ r^{2} = 249.48/2π m^{2}

⇒ r^{2} = 249.48 ÷ (2 × 22/7)

⇒ r^{2} = 39.69

⇒ r = √39.69

⇒ r = 6.3 m

The volume of the air inside the dome will be the same as the volume of the hemisphere.

Now the volume of the air inside the dome = 2/3πr^{3}

= 2/3 × 22/7 × 6.3 m × 6.3 m × 6.3 m

= 523.9 m^{3} (approx.)

Therefore, the inner surface area of the dome is 249.48 m^{2} and he volume of the air inside the dome is 523.9 m^{3}.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 13

**Video Solution:**

## A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹4989.60. If the cost of white-washing is ₹20 per square meter, find the i) inside surface area of the dome, ii) volume of the air inside the dome

NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.8 Question 8

**Summary:**

It is given that dome of a building is in the form of a hemisphere. From inside, it was whitewashed at the cost of ₹4989.60. We have found that the inner surface area of the dome is 249.48 m^{2} and the volume of the air inside the dome is 523.9 m^{3}.

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