Question

A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.

Answer 1

Given base diameter of cone$d_1$ = 20cm

Radius $r_1=\frac{20}{2}=10cm$

Top diameter of Cone$d_2=12cm$

Radius $r_2=\frac{12}{2}=6cm$

Height of the cone (h)= 3cm

Volume of the frustum right circular cone

$=\frac{\pi }{3}(r_1^2+r_2^2+r_1{r}_2)h$

$=\frac{\pi }{3}({10}^2+6^2+10\times 6)3=616c{m}^3$

Let ‘L’ be the slant height of cone

$L=\sqrt{(r_1-r_2{)}^2+h^2}$

$=\sqrt{(10-6{)}^2+3^2}$

$=\sqrt{25}=5cm$

∴∴ Slant height of cone L= 5 cm

Total surface area of the cone =$=\pi (r_1+r_2)\times L+\pi \times r_1^2+\pi \times r_2^2$

$=\pi (10+6)\times 5+\pi \times {10}^2+\pi \times 6^2$

$=\pi (80+100+36)=\pi \times 216=678.85c{m}^2$

∴∴

Total surface area of the cone = 678.85 cm2