Derive the formula for the volume of the frustum of a cone.
Now, approach the question in the same way as the previous one and prove that
ΔABG ~ ΔADF
\(\frac{DF}{BG}\) = \(\frac{AF}{AG}\) = \(\frac{AD}{AB}\)
\(\frac{r_2}{r_1}\) = \(\frac{h_1 - h}{h_1}\) = \(\frac{l_1 - l}{l_1}\)
Now, rearrange them in terms of h and h1
\(\frac{r_2}{r_1}\) = \(1 - \frac{h}{h_1}\) = \(1 - \frac{l - l_1}{r_1}\)
\(1 - \frac{h}{h_1}\) = \(\frac{r_2}{r_1}\)
\(\frac{h}{h_1}\)=\(1 -\frac{r_2}{r_1}\) = \(\frac{r_1 - r_2}{r_1}\)
\(\frac{h_1}{h}\) = \(\frac{r_1}{r_1 - r_2}\)
\(h_1 = \frac{r_1h}{r_1 - r_2}\)
\(l_1 = \frac{r_1 l}{r_1 - r_2}\)
The total volume of frustum of the cone will be = Volume of cone ABC – Volume of cone ADE
= (1/3)πr12h1 - (1/3)πr22(h1 – h)
= (π/3)[r12h1 - r22(h1 – h)]
= \(\frac{\pi}{3} \big[ r_1^2(\frac{hr_1}{r_1 - r_2}) - r_2^2(\frac{hr_1}{r_1 - r_2} - h) \big] \)
= \(\frac{\pi}{3} \big[ (\frac{hr_1^3}{r_1 - r_2}) - r_2^2(\frac{hr_1 - hr_1 + hr_2}{r_1 - r_2}) \big] \)
= \(\frac{\pi}{3} \big[ \frac{hr_1^3}{r_1 - r_2} - \frac{hr_2^3}{r_1 - r_2} \big] \)
= \(\frac{\pi}{3}h \big[ \frac{r_1^3 - r_2^3} {r_1 - r_2} \big] \)
= \(\frac{\pi}{3}h \big[ \frac{(r_1 - r_2)(r_1^2 + r_2^2 + r_1r_2)}{r_1 - r_2} \big] \)
Now, solving this we get,
∴ Volume of frustum of the cone = (1/3)πh (r12 + r22 + r1r2)