Question

A cone of radius 10 cm and height 16 cm is divided into two parts by a plane through the midpoint of its axis parallel to its base. Find the ratio of the volumes of the two parts.A cone of radius 10 cm and height 16 cm is divided into two parts by a plane through the midpoint of its axis parallel to its base. Find the ratio of the volumes of the two parts.​

Answer 1

Step-by-step explanation:

Let the height of the given cone =h cm

On dividing it into two parts, we get

(1) Frustum of the cone with radius R=10 cm and radius = 5cm, and height = h/2 cm

(ii) a smaller cone of radius=5cm and height =h/2 cm

Ratio of the volumes =

volume of the smaller cone

volume of the frustum of the cone

$\frac{\frac{1}{3}\pi \: r {}^{2} ( \frac{h}{2} )}{ \frac{1}{3} \pi( \frac{h}{2}) } \: r1 {}^{2} + r2 {}^{2} + r1 \times r2$

$= \frac{5 \times 5}{10 {}^{2} + 5 {}^{2} + 10 \times 5 }$

$= \frac{25}{175}$

$= \frac{1}{7}$

volume of the smaller cone : volume of the frustum = 1:7

HOPE THIS HELPS YOU

PLEASE MARK AS BRAINLIEST

Answer 2

Given:

The radius of the cone = 10cm

Height of the cone = 16cm

To Find:

The ratio of the volumes of the cone.

Solution:

ΔAGD = ΔAHC  [AAA similarity criteria]

AG = GH [height of the cone]

$\frac{AG}{AH} = \frac{GD}{HC}$ [ corresponding sides of similar triangles]

$\frac{h}{2h} = \frac{GD}{10}$

2GD = 10

GD = 5cm.

Then,

The volume of the cone AED

⇒ 1/3πr²h

⇒1/3π×(5)²×h

⇒1/3π×25×h.. (i)

Now,

volume of the frustum BCDE of the cone = volume of cone ABC - the volume of cone AED

⇒ 1/3π×100×2h - 1/3π×25×h

⇒ 1/3×π×h(200-25)

⇒ 1/3×π×h(175)..(ii)

So, the ratio of the two parts

Dividing (i) and (ii),

⇒ (1/3×π×25×h)/(1/3×π×h×175)

⇒ 25/175

⇒1/7

⇒1:7

Therefore, the ratio of the two parts is 1:7.