# A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

**Solution:**

Given: A right circular cone (joker’s cap) of base radius 7 cm and height 24 cm.

Since the cap is conical in shape, the area of the sheet required to make each cap will be equal to the curved surface area of the cone.

The curved surface area of a right circular cone of base radius(r) and slant height(l) is πrl

Slant height, l = √r² + h² where h is the height of the cone.

Radius, r = 7 cm Height, h = 24 cm

Slant height,

l = √r² + h²

= √(7)² + (24)²

= √49 + 576

= √625

= 25 cm

Area of the sheet required to make each cap = πrl

= 22/7 × 7 cm × 25 cm

= 550 cm²

Area of the sheet required to make 10 such caps = 10 × 550 cm^{2} = 5500 cm^{2}

Thus, the area of the sheet required to make 10 such caps is 5500 cm^{2}.

**Video Solution:**

## A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

### Class 9 Maths NCERT Solutions - Chapter 13 Exercise 13.3 Question 7:

**Summary:**

It is given that there is a joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. We have found that the area of the sheet required to make 10 such caps is 5500 cm^{2}.