# Mensuration-6

### From Homeworkwiki

**A circular tent is in the form of a cylinder surmounted by a cone. If the common diameter**

**is 42 m, the height of the cylindrical portion be 6 m and the highest point of the roof 34 m**

**from the ground, find in square meters, the area of the canvas used in making the tent.**

**Solution:** The curved surface area of the cylindrical portion = 2 π r h

= 2 x 22/7 x 21 x 6

= 792 sq. m

The height of the conical portion = 34 - 6 = 28 m

Therefore, the slant height of the cone = √(h^{2} + r^{2})

= √(28^{2} + 21^{2})

Hence, l = √(784 + 441) = √1225 = 35 m

The surface area of the conical portion of the tent = π r l

= 22/7 x 21 x 35 = 2310 sq. m

The area of the canvas used in making the tent = (2310 + 792) sq. m

= 3102 sq. m