# Tilted tank

Can you find the height of the water in this tilted tank when it is flat?

A fish tank is a cuboid, 100 cm by 60 cm by 40 cm. It is partly filled with water, so that when it is resting on a 60 cm edge, the water level reaches the top edge and the midpoint of the base, as shown.

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What will the height of the water be when the tank is resting on its base (a 100 cm by 60 cm face)?

*This problem is adapted from the World Mathematics Championships*

**Finding the volume of water in the tank**

The water in the tank is in the shape of a triangular prism, as shown, so its volume can be found using length $\times$ area of cross-section.

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The cross-section is the triangle on the original diagram, shown again below with the sides labelled.

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Because the two sides which are labelled are perpendicular, they can be considered to be the base and height of the triangle, so its area is $\frac{1}{2}\times40\times50=1000$ cm$^2$. So the volume of water in the tank is $60\times1000=60\hspace{1mm}000$ cm$^3$.

When the tank is resting on its base, as shown below, the volume of water in the tank is given by $100\times60h=6000h$ cm$^3$.

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But the volume of water in the tank must also still be $60\hspace{1mm}000$ cm$^3$, so $6000h=60\hspace{1mm}000$. So $h$ must be 10.

So the height of the water is 10 cm.

**Using area only**

Because the 60 cm edge will stay in contact with the ground as the tank is rotated, the water in the tank will always form a prism whose length is along the 60 cm edge. So since the volume remains unchanged, the area of the cross-section as seen through the 40 cm by 100 cm face will also remain unchanged.

__Splitting the face into triangles__

When the tank is tilted, that area is the area of the triangle shown below.

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The triangle reaches the midpoint of the longer side, so reflecting in the line of symmetry shown below will create 4 congruent triangles inside the rectangle.

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So the tank must be $\frac{1}{4}$ full.

So when it is flat, the height of the water will be $\frac{1}{4}$ of the height of the tank, which is $\frac{1}{4}$ of 40 cm, which is 10 cm.

__Finding the area of the water__

When the tank is tilted, that area is the area of the triangle shown below.

Image

Because the two sides which are labelled are perpendicular, they can be considered to be the base and height of the triangle, so its area is $\frac{1}{2}\times40\times50=1000$ cm$^2$.

When the tank is flat, that area is the area of the rectangle shown below.

Image

The area of the rectangle is $100\times h$ cm$^2$, so, since this area is still the same as the area of the triangle, $100h=1000$, so $h=10$.

So the height of the water is 10 cm.