Question

If the right circular cone is cut into three solids of volumes $V_1,V_{2}and{V}_3$ by two cuts which are parallel to the base and trisects the altitude, then $V_1:V_2:V_3$ is:

• A. 1:2:3
• B. 1:4:6
• C. 1:6:9
• D. None of these

Answer 1

The correct option is D None of these

Solution: -

The resultant figure is made of three similar triangles. The height and radius will be in ratio $\frac{1}{3}:\frac{2}{3}:1$ = 1:2:3.

r${}_1$ = 1, h${}_1$ = 1

r${}_2$ = 2, h${}_2$ = 2

r${}_3$ = 3, h${}_3$ = 3

Volume will be in the ratio of r\(^2\)h for the three circular cones.

Required Volumes are

V${}_1$ = r${}_1^2$ x h${}_1$ = 1

V${}_2$= r${}_2^2$ x h${}_2$ - V${}_1$ = 8 - 1 = 7

V${}_3$= r${}_3^2$ x h${}_3$ - (V${}_1$ + V${}_2$) = 27 - (7 + 1) = 19

Volumes will be in given ratio = 1:7:19. Option (d).