5. Two right-angled triangles are similar. The larger triangle has short sides which are $1 \mathrm{~cm}$ and $5 \mathrm{~cm}$ longer than the short sides of the smaller triangle. The area of the larger triangle is $8 \mathrm{~cm}^{2}$ more than the area of the smaller triangle. Find all possible values for the side lengths of the short sides of the smaller triangle.
Solution
Let the side lengths of the smaller triangle be $x$ and $y$.
There are two cases to consider: the corresponding sides are the same way round or the other way round.

Case 1:
The first case gives the equation for similarity using ratio of sides as $\frac{x}{y}=\frac{x+1}{y+5}$.
The condition on the area gives $\frac{1}{2} x y+8=\frac{1}{2}(x+1)(y+5)$.
The first equation becomes $y=5 x$ and this can be substituted into the second equation.
This gives $5 x^{2}+16=5 x^{2}+10 x+5$ which can be solved to give $x=1.1$ and $y=5.5$. This is a solution to the problem.
Case 2:
The second case gives the equation for similarity using ratio of sides as $\frac{\tilde{y}}{y}=\frac{y+\nu}{x+1}$.
The condition on the area still gives $\frac{1}{2} x y+8=\frac{1}{2}(x+1)(y+5)$.
The second equation becomes $y+5 x=11$ and this can be substituted into the first equation.
This gives $x(x+1)=(11-5 x)(16-5 x)$ which can be simplified to $24 x^{2}-136 x+176=0$. This can be solved to give $x=2$ and $y=1$ or $x=\frac{11}{3}$ and $y=\frac{-22}{3}$. Since solutions cannot be negative, only the first pair is a solution to the problem.
The solutions are $(x, y)=(1.1,5.5)$ or $(2,1)$