Check sibling questions

Ex 13.8, 8 - A dome of a building is in form of hemisphere - Volume Of Hemisphere

Ex 13.8, 8 - Chapter 13 Class 9 Surface Areas and Volumes - Part 2

Ex 13.8, 8 - Chapter 13 Class 9 Surface Areas and Volumes - Part 3 Ex 13.8, 8 - Chapter 13 Class 9 Surface Areas and Volumes - Part 4

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 11.4,8 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs 498.96. If the cost of white-washing is Rs 2.00 per square metre, find the inside surface area of the dome. We know that Area whitewashed × Cost of white wash = Total cost Area whitewashed × 2 = Rs 498.96 Area whitewashed = 498.96/2 Area whitewashed = 249.48 m2 Now, area whitewashed is curved surface area of hemisphere (as only walls are whitewashed, not floor) Therefore, inner surface area of dome = 249.48 m2 Ex 11.4,8 (ii) volume of the air inside the dome. Volume of air inside dome = Volume of hemisphere = 2/3 πr3 Let the radius of dome = r m First we find radius using surface area Surface area of dome = 249.48 m2 2πr2 = 249.48 2 × 22/7 × r2 = 249.48 r2 = (249.48 × 7)/(2 × 22) r2 = 39.69 r = √39.69 ∴ r = 6.3 m Volume of the air inside the dome = volume of hemisphere = 2/3 πr3 = 2/(3 ) ×22/7 × 6.3 × 6.3 ×6.3 m3 = 523.908 m3 (approx.)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.