# Mensuration-10

### From Homeworkwiki

**A tent is in the form of a cylinder surmounted by a cone. Its diameter is 10.5 m.**

**The height of the cylindrical portion is 4 m and the greatest height**

of the tent is 6.5 m.

of the tent is 6.5 m.

**Find the volume of air in cubic meters. What is the area of**

the canvas required to make the tent?

the canvas required to make the tent?

**Solution:**Radius of the cylinder = 10.5/2 = 5.25 m

Volume of the cylinder = π r^{2} h

= 22/7 x 5.25 x 5.25 x 4 cu.m

= 22/7 x 5.25 x 21 = 346.5 cu.m

Volume of the conical part

= π r^{2} h/3 = 1/3 x 22/7 x 5.25 x 5.25 x 2.5

= 72.1875 cu.m

Volume of the air in the tent = volume of the conical part + volume of the cylindrical part

= (346.5 + 72.1875) cu. m

= 418.6875 cu. m

Slant height of the conical part = l = √(h^{2} + r^{2})

= π(2.5^{2} + 5.25^{2}) = π(6.25 + 27.5625)

= π(33.8125) = 5.81 (approximately)

Curved surface area of the cone = π r l = 22/7 x 5.25 x 5.81 sq.m

= 95.865 sq. m

Curved surface area of the cylindrical part = 2 π r h

= 2 x 22/7 x 5.25 x 4 = 132 sq. m

Area of canvas required = 132 + 95.865 sq. m = 227.865 sq. m