Question

A hollow cone is cut by a plane parallel to the base and the upper portion is removed. If the curved surface area of the remainder is $\frac{8}{9}^{th}$ of the curved surface of the whole cone, the ratio of the line segments into which the cone's altitude is divided by the plane is given by

• A. 2:3
• B. 1:3
• C. 1:2
• D. 1:4

Answer 1

Answer: (C)

1:2

Answer 2

Let R and r be the radius and $l$1 and $l$2 be slant height
of bigger and smaller cone respectively.
Curved surface area of cone = $\pi$R$($ $l$1 + $l$2 $)$
Curved surface Area of remainder = \pi$$(R + r$)$ $l$2
According to question
$\frac{8}{9}$×$\pi$R$($ $l$1 + $l$2 $)$=\pi$$(R + r$)$ $l$2
8R $l$1= $l$2 $($R + 9r$)$
\frac{l_1}{l_2}=\frac{R + 9r}{8R}=$$($$\frac{1}{8}$$+$$\frac{9r}{8R}$$)...........(i)
According to sin rule =$\frac{R}{l_1 + l_2}=\frac{r}{l_1}$
$\frac{r}{R} = \frac{l_1}{l_1 + l_2}$.........from eq(i)

\frac{l_1}{l_2}=($$\frac{1}{8} + \frac{9}{8}×\frac{l_1}{l_1 + l_2}$$)

⇒$\frac{8l_1}{l_2} - \frac{9l_1}{l_1 + l_2}=1$
$8l_1^2 + 8l_1l_2-9l_1l_2=l_1l_2+l_2^2$
$8l_1^2 - 4l_1l_2+2l_1l_2-l_2^2$
$4l_1(2l_1-l_2)+l_2(2l_1-l_2)=0$
$(2l_1-l_2)(4l_1 + l_2)=0$
$2l_1=l_2$
$\frac{l_1}{l_2}=\frac{1}{2}$
So the correct option is 1:2