Question

∆ABC IS isosceles triangle with AB=AC=b cm,BC=a cm and AD is perpendicular to BC is made to revolve about AD. Write a formula to the height of the cone so formed in terms of a and b

Answer 1

$\sqrt{b {}^{2} - a {}^{2} }$

Answer 2

Answer:

$\textbf{The height of cone formed by revolving = }\bf\frac{1}{2}\times \sqrt{4b^2-a^2}$

Step-by-step explanation:

Given : AB = AC = b cm and BC = a cm

When the triangle is revolved about AD then the figure which is formed is of cone.

$\text{Now, area of base of cone whose radius is }\frac{a}{2}\text{ is given by :}\\\\Area=\pi\times r^2\\\\Area=\pi\times(\frac{a}{2})^2$

In right angled triangle formed at the base of cone, height can be found by using Pythagoras theorem :

$h^2=b^2-r^2\\\\h^2=b^2-\frac{a^2}{4}\\\\h^2=\frac{4b^2-a^2}{4}\\\\h=\sqrt{\frac{4b^2-a^2}{4}}\\\\\implies h=\frac{\sqrt{4b^2-a^2}}{2}\\\\\textbf{Hence, The height of cone formed by revolving = }\bf\frac{1}{2}\times \sqrt{4b^2-a^2}$