Question

The height of a solid cone is $12cm$ and the area of the circular base is $64\pi c{m}^2$.A plane parallel to the base of the cone cuts through the cone $9cm$ above the vertex of the cone, the area of the base of the new cone so formed is

• A. $9\pi c{m}^2$
• B. $16\pi c{m}^2$
• C. $25\pi c{m}^2$
• D. $36\pi c{m}^2$

Answer 1

Answer: (D)

Height of a solid cone $\left(h\right)=12cm$

Area of circular base= $64\pi c{m}^2$

Area of base (circle) $=\pi (r_1{)}^2$

$\Rightarrow \text{Radius}\left(r_1\right)=\sqrt{\frac{\text{Area}}{\pi }}=\sqrt{\frac{64\pi }{\pi }}cm=\sqrt{64}=8cm$

$\therefore r_1=8cm$

In $△OABand△OCD$

In $△OAB\text{and}△OCD$

$\angle O=\angle O\left(\text{Common}\right)\angle OAB=\angle OCD\left(\text{Each}{90}^{\circ }\right)$

$\Rightarrow △OAB\sim △OCD$ [Since, By AA similarity criterion]

$\Rightarrow \frac{OA}{OC}=\frac{AB}{CD}$

$\Rightarrow \frac{12}{9}=\frac{8}{CD}$

$\Rightarrow CD=\frac{9\times 8}{12}$

$\Rightarrow CD=6cm$

Now in cone $OCD$,

Radius $\left(r_2\right)=6cm$

$\therefore$ Base area $=\pi \left(\right(r_2{)}^2$

$=\pi \times 6cm\times 6cm$

$=36\pi c{m}^2$

Therefore, the base area of new cone is $36\pi c{m}^2$

Hence, Option D is correct.