Troy has a container in the shape of a rectangular prism with a \(40\text{ cm}\) by \(60\text{ cm}\) base and a height of \(30\text{ cm}\). He labels the vertices of a \(40\text{ cm}\) by \(30\text{ cm}\) side face \(A\), \(B\), \(C\), and \(D\), with \(A\) and \(B\) being vertices of the base face too. He then puts some water in the container and tilts the container along \(AB\) until the water completely covers face \(ABCD\). (He is able to do this so that no water is lost!) At this point, the water still covers \(\frac{4}{5}\) of the base area.
Determine the depth of the water, in centimetres, when the container is level.
Let the other vertex on the bottom face adjacent to \(A\) be labelled \(E\). Thus, \(EA=60\text{ cm}\). Let \(P\) be the point on \(EA\) such that \(AP=\frac{4}{5}(EA)=\frac{4}{5}(60)=48\text{ cm}.\)
When the tank is tilted so that the water completely covers side face \(ABCD\), a triangular prism with triangular base \(ADP\) and height \(40\text{ cm}\) is created. Also, \(\triangle ADP\) is a right-angled triangle, so when finding the area of \(\triangle ADP\) we can use \(AP\) as the base of the triangle and \(AD\) as the height of the triangle. That is, \[\begin{aligned} \text{Volume of triangular prism}&=\text{Area of } \triangle APD \times \text{height of triangular prism}\\ &=\frac{1}{2} (AP)(AD) \times (AB)\\ &=\frac{1}{2} (48)(30) \times (40)\\ &=28\,800 \text{ cm}^3 \end{aligned}\] Let \(h\) represent the height of the water when the tank is level. The volume of the rectangular prism with base \(40\text{ cm}\) by \(60\text{ cm}\) and height \(h\) is the same as the volume of the triangular prism formed when the tank is tilted. That is, \[\begin{aligned} 60 \times 40 \times h &=28\,800\\ 2400h &=28\,800\\ h &=12 \end{aligned}\] Therefore, the water is \(12\text{ cm}\) deep when the container is level.