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

The Correct Option is C

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