A cone of radius 4 cm is divided into two parts by drawing a plane through the mid-point of its - Maths - Surface Areas and Volumes

Question

A cone of radius 4 cm is divided into two parts by drawing a
plane through the mid-point of its axisand parallel to its base
Compare the volumes of the two parts

Lalit Mehra · Accepted Answer

Let the height of the given cone be H cm.

Radius of the cone = 4 cm

The cone is divided into two parts by drawing a plane through the mid points of its axis and parallel to the base. One part is a smaller cone and the other part is a frustum of cone.

$∴ OC = CA = \frac{H}{2} cm$

Let the radius of the smaller cone be r cm.

In ∆OCD and ∆OAB,

∠OCD = ∠OAB (90°)

∠COD = ∠AOB (Common)

∴∆OCD ∼ ∆OAB (AA Similarly criterion)

$\Rightarrow \frac{OA}{OC} = \frac{AB}{CD} = \frac{OB}{OD} (Corresponding sides are proportional) ∴ \frac{H}{\frac{H}{2}} = \frac{4 cm}{r}$

⇒ 2r = 4 cm

⇒ r = 2 cm

$Volume of smaller cone = \frac{1}{3} π (CD)^{2} \times OC = \frac{1}{3} π (2 cm)^{2} \times \frac{H}{2} cm = \frac{2 π H}{3} cm Height of frustum of cone = \frac{H}{2}$

Volume of frustum of cone

$= \frac{1}{3} π (\frac{H}{2}) [(4 cm)^{2} + (2 cm)^{2} + 4 cm \times 2 cm] {∵ Volume of frustum = \frac{1}{3} π h ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2})} = \frac{π H}{6} (16 + 4 + 8) {cm}^{3} = \frac{28 π H}{6} {cm}^{3} = \frac{14 π H}{3} {cm}^{3} \frac{Volume of smaller cone}{Volume of frustum of cone} = \frac{\frac{2 π H}{3} {cm}^{3}}{\frac{14 π H}{3} {cm}^{3}} = \frac{1}{7} ∴ Volume of the smaller cone = \frac{1}{7} \times Volume of frustum of cone$

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