Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Examples

Example 1

Example 2 Important

Example 3

Example 4

Example 5

Example 6 Important

Example 7

Example 8 Important

Example 9

Example 10

Example 11 Important

Example 12

Example 13

Example 14 Important

Example 15

Example 16 Important You are here

Example 17

Example 18 Important

Example 19

Surface Area and Volume Formulas Important

Chapter 13 Class 9 Surface Areas and Volumes

Serial order wise

Last updated at Aug. 26, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Example 16 Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m2, find the volume of the tent that can be made with it. Let tent have radius = r = 7m Let slant height = l m Let height = h m Area of canvas = Area available for making tent + Wastage 551 m2 = Area available for making tent + 1 m2 Area available for making tent = 551 m2 – 1 m2 = 550 m2 Area available for making tent = Curved surface area of conical tent 550 = πrl 550 = 22/7 × 7 × l 550 = 22 × 1 × l 22 × 1 × l = 550 l = 550/22 m l = 25 m We know that l2 = h2 + r2 252 = h2 + 72 252 – 72 = h2 h2 = 252 – 72 h2 = (25 – 7) (25 + 7) h2 = (18) (32) h = √("18(32)" ) h = √("(9 × 2) × (32)" ) h = √("(9) × (64)" ) h = √("(32) × (82)" ) h = 3 × 8 h = 24 m Volume of tent = 1/3πr2h = 1/3 × 22/7 × 7 × 7 × 24 m3 = 22 × 1 × 7 × 8 m3 = 1232 m3.