Question

Two cones have their 1:3 and the radie of heights in the ratio their bares in show that their volumes are ratio 3:1 . are the in the ratio 3.1.​

Answer 1

Answer:

Let the heights be 1x and 3x respectively

and their radii be 3y and 1y .

volume of cone1 = volume of cone 2

1/3 \pi r ^{2} h =1/3 \pi r ^{2} h

1/3 \pi (3y) ^{2} x =1/3 \pi (1y) ^{2} 3x

9y ^{2}  * x =y ^{2} *3x

9:3

3:1

Step-by-step explanation:

Answer 2

Given:-

Two cones have their 1:3 and the radie of heights in the ratio their bares in show that their volumes are ratio 3:1 . are the in the ratio 3.1.

Need To Find:-

Their volumes

Solution:-

Let the height of two cones be h, 3h and their radii be 3r, r respectively.

$\small\sf{∵ Volume \: of \: cone = \frac{1}{r} \:\pi R²} \: H$

$\small\sf{∴ V_{1}} \frac{1}{3} \pi {}^{} (3r) {}^{2} h = 3\pi {}^{2} h$

$\small\sf{and \: { V_{2}} = \frac{1}{3} \pi {}^{2} (3h) = \pi {}^{2} h}$

∴ Ratio of volumes of the cones,

$\small\sf \frac{ V_{1}}{ V_{2} } = \frac{3\pi {}^{2}h }{\pi {}^{2}h } = \frac{3}{1} = 3:1$

Hence Proved