Question

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

Answer 1

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

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

Answer 2

Answer: 84yd

Step-by-step explanation:

Vcylinder=πr2h

=π(2)2(6)

=24π

≈24(3.14)

≈75.36 yd

_____________________

The height of the cone is the height of the figure minus the height of the cylinder:

8 yd−6 yd=2 yd

Vcone=13πr2h

=13π(2)22

=223π

≈223(3.14)

≈8.37 yd3

Add the volumes.

75.36 yd3+8.37 yd3=83.73 yd3

To the nearest cubic yard, the volume is 84 yd3.