If the radius of the base of a cone is doubled, and the volume of the new cone is three times the volume of the original cone, then what will be the ratio of the height of the original cone to that of the new cone?
Volume of the cone = $$\frac{1}{3} \times \pi r^2 h$$
$$\frac{v_1}{v_2} = \frac{(r_1)^2(h_1)}{(r_2)^2(h_2)}$$
$$\frac{h_1}{h_2} = \frac{(r_1)^2(v_2)}{(r_2)^2(v_1)}$$
$$\frac{h_1}{h_2} = \frac{(2r_1)^2(v_2)}{(r_2)^2(3v_2)}$$Â
$$h_1 :Â h_2 = 4 :Â 3$$
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