1. ## Rectangular Container

A rectangular container has a base of 10 by 6 cm. The sides are made to slope outward so that the upper edges are 12 by 8 cm.Find the volume if the height is 5cm. If it is filled with water of altitude 3cm, what is the volume of water?

2. Hello, reiward!

A container has a base of 10 by 6 cm.
The sides are sloped outward so that the upper edges are 12 by 8 cm.

(a) Find the volume if the height is 5cm.

(b) If it is filled with water of altitude 3cm, what is the volume of water?
We are dealing with the volume of a frustum of a pyramid.
. . Lucky for both of us, there is a formula for this!

Formula: . $V \;=\;\frac{h}{3}\left(B_1 + \sqrt{B_1B_2} + B_2\right)$ . . where: . $\begin{Bmatrix} h &=& \text{height} \\ B_1 &=& \text{area of one base} \\ B_2 &=& \text{area of other base} \end{Bmatrix}$

(a) We have: . $\begin{array}{ccc}h &=& 5 \\ B_1 &=& 60 \\ B_2 &=& 96 \end{array}$

. . Therefore: . $V \;=\;\frac{5}{3}\bigg(60 + \sqrt{60\cdot96} + 96\bigg) \;=\;\frac{5}{3}\left(156 + \sqrt{5760}\right)$

. . . . . . . . . . $V \;=\;\frac{5}{3}\left(156 + 24\sqrt{10}\right) \;=\;20\left(13+2\sqrt{10}\right)\text{ cm}^3$

(b) This time, $h = 3$

The bottom area is still: . $B_1 \,=\,60$

And we need the area of the top surface of the water.
After some geometric acrobatics, it is : . $B_2 \:=\:7.2 \times 11.2 \:=\:80.64$

Therefore: . $V \;=\;\frac{3}{3}\left(60 + \sqrt{(60)(80.64)} + 80.64\right) \;=\;140.64 + \sqrt{4838.4}$

. . . . . . . . . . $V \;=\;140.64 + 48\sqrt{2.1} \;=\;48\left( 2.93 + \sqrt{2.1}\right)\text{ cm}^3$