Following on from the earlier Cake problem...
Solution
Focus on the triangle with base length \(y\). It has a perpendicular height of \(\frac{1}{2}(x+y)\) therefore an area of \(\frac{y(x+y)}{4}\).
The entire cake has an area of \(\frac{3}{4}(x+y)^2\) so this triangle is a quarter of that.
\(\frac{y(x+y)}{4}=\frac{1}{4} \times \frac{3}{4}(x+y)^2\)
\(y(x+y)=\frac{3}{4}(x+y)^2\)
\(y=3x\)
So the ratio is 1:3.
Given that Pythagoras' Theorem involves the sum of squares, it's common to see squares arranged around the sides of a right angled triangle.
Here are some questions where other shapes are considered.
The semicircular problem
By Pythagoras' Theorem, the hypotenuse is 20cm. The areas of the semicircles are \(\frac{1}{2}\pi\times 6^2=18\pi\), \(\frac{1}{2}\pi\times 8^2=32\pi\) and \(\frac{1}{2}\pi\times 10^2=50\pi\). The sum of the areas of the two smaller semicircles is the area of the larger semicircle.
The equilateral triangle problem
The hypotenuse is 13cm. The area of an equilateral triangle is \(\frac{\sqrt{3}}{2}x^2\) where \(x\) is the side length. The areas of the triangles are therefore \(\frac{\sqrt{3}}{2}5^2=\frac{25\sqrt{3}}{2}\), \(\frac{\sqrt{3}}{2}12^2=72\sqrt{3}\) and \(\frac{\sqrt{3}}{2}13^2=\frac{169\sqrt{3}}{2}\).
Like before, the sum of the areas of two smaller triangles is the area of the larger triangle.
The similar shape problem
The hypotenuse is 50cm. Shape Q is similar to shape P with a linear scale factor of \(\frac{4}{5}\). This means that the area scale factor is \(\frac{16}{25}\). The area of Q is then \(\frac{16}{25}\times 1800=1152 cm^2\)