Question

The figure shows a vertical cross section of a sphere inscribed in a right cone

Answer 1

Answer:

C)49π

Step-by-step explanation:

Given:

As in given figure there are two right triangle,

Let the bigger right triangle be  ABC

and the smaller right angle be ADE

Finding length of hypotenuse of ΔABC

h $=\sqrt{12^{2} +5^{2} }\\ =13$

h=AC=13

Now as the two right triangles are similar as both have one angle common i.e. A, and one right angle i.e. 90 degree.

As ΔABC is similar to ΔADE

BC/AC=DE/AE

5/13=r/(12-r)

r= (12-r)0.384

r= 4.615-0.384r

r+0.384r =4.615

1.384r=4.615

r=4.615/1.384

r=3.33

Now volume of the inscribed sphere

volume=4/3πr^3

putting value of r=3.33 in above

volume=4/3π(3.33^3)

           =49.23π

            = 49π !