Question

A hollow cone is cut by a plane parallel to the base and upper part is removed. If the curved surface of the remainder is $\frac{15}{16}$ of the curved surface of the whole cone, find the ratio of the line-seaments into which the cones altitude is divided by the plane.

• A. $1:3$
• B. $2:3$
• C. $1:1$
• D. $5:1$

Answer 1

The correct option is A $1:3$

Let $R,H,L$ be the radius, height and slant height of the original cone respectively

and $r,h,l$ be the radius, height and slant height of the smaller cone respectively.
In $△OAB$ and $△OCD$
$\angle OAB=\angle OCD$ ....$\left({90}^0\right)$
$\angle AOB=\angle COD$ ....(Common)
$\therefore △OAB\cong △OCD$ (AA test of similarity)

$\Rightarrow \frac{OB}{OD}=\frac{AB}{CD}=\frac{OA}{OC}$ ...(Corresponding sides are proportional)

$\Rightarrow \frac{l}{L}=\frac{r}{R}=hH$ ....(i)

Curved surface area of the smaller cone=Curved surface area of cone-Curved surface area of the frustrum

=$1-\frac{15}{16}\times \text{Curved surface area of cone}$

=$\frac{1}{16}\times \text{Curved surface area of cone}$

$\therefore \frac{\text{Curved surface area of the smaller cone}}{\text{Curved surface area of cone}}=\frac{1}{16}$

$\Rightarrow \frac{\pi rl}{\pi RL}=\frac{1}{16}$

$=>\frac{r}{R}\times \frac{l}{L}=\frac{1}{16}$

$=>\frac{h}{H}\times \frac{h}{H}=\frac{1}{16}$ ...Using (i)

$=>(\frac{h}{H}{)}^2=(\frac{1}{4}{)}^2$

$=>\frac{h}{H}=\frac{1}{4}$

$=>h=\frac{H}{4}$

$\frac{OA}{AC}=\frac{h}{H-h}$

$=\frac{\frac{H}{4}}{H-\frac{H}{4}}$

$=\frac{\frac{H}{4}}{\frac{3H}{4}}$

$=\frac{1}{3}$

Thus, the cones altitude is divided in the ratio=$\frac{1}{3}$