Question

The radius of the base of a right circular cone is 'r'. It is cut by a plane parallel to the base at a height 'h' from the base. The distance of the boundry of the upper surface from the centre of the base of the frustum is whole underoot h2+rr/9. Show that the volume of the frustum is 13/27Pier2h.

Answer 1

Answer:

V = $\frac{12πr^{2}h }{27}$

Step-by-step explanation:

Since the radius of the base of a right circular cone is 'r'. It is cut by a plane parallel to the base at a height 'h' from the base.

We want to show that the volume of the frustum is 13/27Pier2h.

Thus we have that:

BD = $\sqrt{h^{2} + \frac{r^{2} }{9}  }$

Then we have that:

ED = $\frac{r}{3}$

We know that our volume is equal to:

= $\frac{πh}{3}$($ED^{2}$ + (ED)(BC) + $BC^{2}$)

=$\frac{πh}{3}$[$\frac{r^{2} }{9}$ + $\frac{r^{2} }{3} + r^{2}$]

=$\frac{12πr^{2}h }{27}$          q.e.d.

Hence, we have proven the lemma

Answer 2

Answer:

Step-by-step explanation: