Answer to Question #145384 in Geometry for jk

Question

A right circular cone is inscribed in a cube having a diagonal which
measures 20√3 cm.
1. What is the altitude of the cone?
2. What is the volume of the cone?
3. What is the surface area of the cone?

Answer 1

Let D = diagonal of the cube

d = diagonal of one face of the cube

l = length of one side of a face of the cube

h = the altitude

r = radius of a cone

Using Pythagoras theorem,

$d²=2l² \textsf{ (since one face of the cube is a square)}\\$ Also, using Pythagoras theorem,

$D²=d²+l²\\ (20\sqrt3)²=2l²+l²\\ 1200=3l²\\ l=\sqrt{400}\\ l=20cm$

1) since the right circular cone is inscribed in the cube, the altitude of the cone is equal to the height of the cube which is equal to the length of one side of a face of the cube.

$h=l\\ h=20cm$

2) The base of a cone is a circle and since the circle is inscribed in a square, the radius of the circle is equal to half of the length of one side of the square.

$r=\frac l2\\ r=\frac{20cm}2\\ r=10cm\\ \textsf{the volume of the cone is given by,}\\ v=\frac 13\pi r²h\\ v=\frac 13 \pi (10)²(20)\\ v=2094.4cm³$

3) The surface area of the cone = curved surface area of the cone + the area of the circular base

$\begin{aligned}\textsf{the curved surface area}&=\pi rl\\ \textsf{the area of the circular base}&=\pi r²\\ \textsf{The surface area of the cone}&= πrl+πr²\\&=πr(l+r)\\&=π×10(20+10)\\&=942.5cm²\end{aligned}$