Question

A cone of base radius 4 cm is divided into two parts by drawing a plane through
the mid-point of its height and parallel to its base. Compare the volume of the
two parts.​

Answer 1

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A cone of radius 4 cm is divided into 2 parts by drawing a plane through mid point of its axis parallel to it's base.

We got two parts Top and bottom

Top part is shape of cone whose height is h and radius r

Volume of cone $=\frac{1}{3}\times \pi\times r^2\times h$

Volume of Top part $=\frac{1}{3}\times \pi\times r^2\times h$

Bottom part is shape of frustum whose height is h and top radius r and bottom radius R.

Using similar theorem property, ΔOAD ≈ΔOBC

$\therefore \dfrac{OA}{OB}=\dfrac{AD}{BC}</p><p></p><p> \dfrac{h}{2h}=\dfrac{r}{R}$

R=2r

Volume of bottom part $=\frac{1}{3}\times \pi\times h(R^2+r^2+Rr)$

Volume of bottom part$=\frac{1}{3}\times \pi\times h(4r^2+r^2+2r^2)=\frac{1}{3}\times \pi\times h\times 7r^2$

Now we find the ratio of both volume $\dfrac{\frac{1}{3}\times \pi\times r^2\times h}{\frac{1}{3}\times \pi\times h\times 7r^2}$

Ratio = 1:7

Hence, volume of bottom part is 7 times the volume of top part.

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