Chapter 13 Class 10 Surface Areas and Volumes

Class 10
Important Questions for Exam - Class 10

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

### Transcript

Example 7 A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy.(Take π = 3.14) Now, Volume of toy = Volume of cone + Volume of hemisphere Volume of cone Height of cone = OA = h = 2 cm Diameter of cone = BC = 4 cm So, Radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 "=" 4/2 = 2 cm Now, Volume of cone = 1/3 𝜋𝑟2ℎ = 1/3 𝜋 × (2)2 × (2) = 𝟖𝝅/𝟑 cm3 Volume of hemisphere Diameter of hemisphere = BC = 4 cm So, Radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 "=" 4/2 = 2 cm Volume of hemisphere = 2/3 𝜋𝑟3 = 2/3×π×(2)3 = 2/3 π×2×2×2 = 𝟏𝟔𝝅/𝟑 cm3 Now, Volume of the toy = Volume of cone + Volume of hemisphere = 𝟖𝝅/𝟑+𝟏𝟔𝝅/𝟑 = (8𝜋 + 16𝜋)/3 = 2/3 π × 2 × 2 × 2 = 𝟏𝟔𝝅/𝟑 cm3 Now, Volume of the toy = Volume of cone + Volume of hemisphere = 𝟖𝝅/𝟑+𝟏𝟔𝝅/𝟑 = (8𝜋 + 16𝜋)/3 = 24𝜋/3 = (24 × 3.14)/3 = 8 × 3.14 = 25.12 cm3 Now, we need to find the difference of the volumes of the cylinder and the toy. Volume of cylinder Cylinder circumscribes the toy. Diameter of cylinder = HG = BC = 4 cm Radius = r = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟/2 "=" 4/2 = 2 cm And, Height of cylinder = OA + OP = Height of cone + Radius of hemisphere = 2 + 2 = 4 cm Now, Volume of cylinder = 𝜋𝑟2ℎ = 3.14×(2)2×(4) = 3.14×4×4 = 50.24 Therefore, Difference of the volume = Volume of cylinder – Volume of toy = 50.24 – 25.12 = 25.12 cm3 Hence, difference of the volume is 25.12 cm3

#### 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.