Q:

In the figure. d=4 yd, h=6 yd, and H=8 yd. What is the approximate volume of the figure? Use 3.14 to approximate

Accepted Solution

A:
check the picture below.

so is really just a cylinder with a cone on the side, notice, since the diameter of the base is 4, the radius is half that then.

so we just simply get the volume of the cylinder, and the cone, and sum them up

[tex]\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h\quad \begin{cases} r=radius\\ h=height\\ -----\\ r=2\\ h=6 \end{cases}\implies V=\pi \cdot 2^2\cdot 6\\\\ -------------------------------\\\\ \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\quad \begin{cases} r=radius\\ h=height\\ -----\\ r=2\\ h=2 \end{cases}\implies V=\cfrac{\pi \cdot 2^2\cdot 2}{3}\\\\ -------------------------------\\\\ \stackrel{cylinder's~volume}{24\pi }~~+~~\stackrel{cone's~volume}{\cfrac{8\pi }{3}}[/tex]