Solve this

Question:

Note Use $\pi=\frac{22}{7}$, unless stated otherwise.

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:

Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

$\therefore$ Slant height of the cone, $l=\sqrt{r^{2}+h^{2}}=\sqrt{7^{2}+24^{2}}=\sqrt{49+576}=\sqrt{625}=25 \mathrm{~cm}$

Area of the sheet required to make one cap = Curved surface area of the cone $=\pi r l=\frac{22}{7} \times 7 \times 25=550 \mathrm{~cm}^{2}$

∴ Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm2

Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm2.

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now