Question

if the height if a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of cone​

Answer 1

Step-by-step explanation:

Let h and r be the height and radius of the cone & volume be V.

$\therefore \: v = \frac{1}{3} \pi \: {r}^{2} h.....(1) \\</p><p>v_1= \frac{1}{3} \pi \: {(r-\frac{50}{100}r )}^{2} (h+\frac{200}{100}h) \\</p><p>\therefore \:v_1=\frac{1}{3} \pi \: {(r-0.5r )}^{2} (h+2h) \\</p><p>\therefore \:v_1=\frac{1}{3} \pi \: {(0.5r )}^{2} (3h) \\</p><p>\therefore \:v_1=\frac{1}{3} \pi \: 0.25r^{2} (3h) \\</p><p>\therefore \:v_1=\frac{1}{3} \pi \: 0.75r^{2} h \\</p><p>\therefore \:v_1=0.75\times (\frac{1}{3} \pi \: r^{2} h) \\</p><p>\therefore \:v_1=0.75\times v\\</p><p>\therefore \:v_1=\frac{75}{100}\times v\\\therefore \:v_1=\frac{3}{4}\times v$

So, if the height of a right circular cone is increased by 200% and the radius of the base is reduced by 50%, then the volume of cone will become $\frac{3}{4}$ times of initial volume.