Question

Two right circular cones A and B are made. A having 3 times the radius of B and
having half the volume of A. Find the ratio of height of A and B.

Answer 1

The ratio between the height of A and B would be 2:9.

Explanation:

It is given that A and B are two right circular cones.

The cone A is having 3 times the radius of B. Let the radius of cone A be $r_{1}$ and radius of cone B be $r_{2}$

It can be written as,

$r_{1}=3 r_{2}$

Also, it is given that B is having half the volume of A. Let the volume of cone A be $V_{1}$ and volume of cone B be $V_{2}$

It can be written as,

$V_{2}=\frac{1}{2} V_{1}$

Now, we shall determine the ratio of height of A and B.

The formula for volume of cone is given by

$V=\frac{1}{3} \pi r^{2} h$

The volume of cone A is $V_{1}=\frac{1}{3} \pi r_{1}^{2} h_{1}$

The volume of cone B is $V_{2}=\frac{1}{3} \pi r_{2}^{2} h_{2}$

To find the ratio, let us divide the volume of A and B.

$\frac{V_{1}}{V_{2}}= \frac{\frac{1}{3}r_{1}^{2} h_{1}}{\frac{1}{3}r_{2}^{2} h_{2}}$

Simplifying,

$\frac{V_{1}}{V_{2}}=\frac{r_{1}^{2} h_{1}}{r_{2}^{2} h_{2}}$

Now, substituting the values $r_{1}=3 r_{2}$ and $V_{2}=\frac{1}{2} V_{1}$, we get,

$\frac{V_{1}}{\frac{J}{2} V_{1}}=\frac{9 r_{2}^{2} h_{1}}{r_{2}^{2} h_{2}}$

$2=\frac{9 h_{1}}{h_{2}}$

Dividing both sides by 9, we get,

$\frac{h_{1}}{h_{2}}=\frac{2}{9}$

Thus, the ratio of between the height of A and B would be 2:9.

Learn more...

1. brainly.in/question/12822301

2. brainly.in/question/13158390