A closed container is in the shape of an inverted right pyramid with a rectangular base.
The dimensions of the base are \(2L\) cm by \(L\) cm and the vertical height of the pyramid is \(H\) cm.
The container initially contains some water.
The volume of the water is \(\dfrac1{27}\) of the total capacity of the container.
Find the area of the rectangular surface of water, in terms of \(L\).[2]
The container is now inverted so that the rectangular base is at the bottom. Given that the total height of the pyramid \(H\) is 90 cm, calculate the new height of the water.
Give your answer to 2 decimal places.[3]
A new container is created, where it has a length \(L\), and a vertical height of \(\dfrac12H\).
From the information given, Charles concludes that, “These two containers are similar solids.”
Justify with reason(s) if you agree with Charles’ claim.[1]