Name: Area of Segment
Uploaded: 2019-07-19
Duration: 7 min 50 s
Description: Do you know the result for the area of a segment? Well, this video talks about the derivation of it. So, let's dive in and master the concept in just a few m...

Question

A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base. Prove that the volumes of the parts are in the ratio

1 : 7 : 19

.

Accepted Answer

Given : A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base.

The height of a Cone, $3 h$ is trisected by $2$ planes parallel to the base of the cone at equal distances.

So, the cone is divided into a smaller cone & two frustums of the cone. The height of each is $‘ h ’$ unit.

$∆ ABG ≅ ∆ ACF ≅ ∆ ADE$ (corresponding sides are to be proportional by $AAA$ Angle-Angle-Angle property)

$\frac{AB}{AC} = \frac{h}{2 h} = \frac{1}{2} = \frac{BG}{CF} = \frac{r}{2 r}$

$\frac{AB}{AD} = \frac{h}{3 h} = \frac{1}{3} = \frac{BG}{DE} = \frac{r}{3 r}$

From the above figure Volume of each shape is

Volume of small Cone $ABG = \frac{1}{3} {πr}^{2} h$ ....(i)

Volume of middle frustum $= \frac{1}{3} πh ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2}) = \frac{1}{3} πh ({BG}^{2} + {CF}^{2} + BG \times CF)$

$\Rightarrow \frac{1}{3} πh ({BG}^{2} + {(2 BG)}^{2} + BG \times 2 BG) = \frac{1}{3} πh (r^{2} + {(2 r)}^{2} + r \times 2 r) = \frac{1}{3} πh (r^{2} + 4 r^{2} + 2 r^{2})$

$\Rightarrow \frac{1}{3} πh \times 7 r^{2}$ ....(ii)

Volume of bottom frustum $= \frac{1}{3} πh ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2}) = \frac{1}{3} πh ({CF}^{2} + {DE}^{2} + CF \times DE)$

$\Rightarrow \frac{1}{3} πh ({(2 BG)}^{2} + {(3 BG)}^{2} + 2 BG \times 3 BG) = \frac{1}{3} πh ({(2 r)}^{2} + {(3 r)}^{2} + 2 r \times 3 r) = \frac{1}{3} πh (4 r^{2} + 9 r^{2} + 6 r^{2})$

$\Rightarrow \frac{1}{3} πh \times 19 r^{2}$ ....(iii)

Now ratio of (i), (ii) & (iii)

i.e, Volume of small Cone $ABG$ $:$ Volume of middle frustum $:$ Volume of bottom frustum

$\frac{1}{3} {πr}^{2} h : \frac{1}{3} πh \times 7 r^{2} : \frac{1}{3} πh \times 19 r^{2}$

$\Rightarrow 1 : 7 : 19$

Hence, the volumes of the parts are in the ratio $1 : 7 : 19$ .

A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base. Prove that the volumes of the parts are in the ratio $1 : 7 : 19$ .

Important Questions on Mensuration

Learn and Prepare for any exam you want !

A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base. Prove that the volumes of the parts are in the ratio 1:7:19.

Important Questions on Mensuration

Learn and Prepare for any exam you want !

A cone is cut into three parts by planes through the points of trisection of its altitude and parallel to the base. Prove that the volumes of the parts are in the ratio $1 : 7 : 19$ .