Answer:
Equal volumes
Step-by-step explanation:
Given
[tex]h \to[/tex] height of cone
[tex]d \to[/tex] diameter of cone
[tex]H \to[/tex] height of cylinder
[tex]D \to[/tex] diameter of cylinder
Such that:
[tex]d = 2D[/tex]
[tex]h =\frac{3}{4}H[/tex]
Required
The relationship between the volumes
The volume of a cylinder is:
[tex]V_1 = \pi R^2H[/tex]
Where
[tex]R = 0.5D[/tex]
So:
[tex]V_1 = \pi (0.5D)^2H[/tex]
[tex]V_1 = \pi *0.25*D^2H[/tex]
[tex]V_1 = 0.25\pi D^2H[/tex]
The volume of the cone is:
[tex]V_2 = \frac{1}{3}\pi r^2h[/tex]
Where
[tex]r =0.5d[/tex]
[tex]r = 0.5 * 2D[/tex]
[tex]r = D[/tex]
and
[tex]h =\frac{3}{4}H[/tex]
So, we have:
[tex]V_2 = \frac{1}{3}\pi * D^2 * \frac{3}{4}H[/tex]
[tex]V_2 = \frac{1}{4}\pi * D^2H[/tex]
[tex]V_2 = 0.25\pi * D^2H[/tex]
[tex]V_2 = 0.25\pi D^2H[/tex]
So, we have:
[tex]V_1 = 0.25\pi D^2H[/tex]
[tex]V_2 = 0.25\pi * D^2H[/tex]
[tex]V_1 = V_2[/tex]
