19. Three squares are arranged as shown so that their bases lie on a straight line. Also, the corners $P, Q$ and $R$ lie on a straight line. The middle square has sides that are $8 \mathrm{~cm}$ longer than the sides of the smallest square. The largest square has sides of length $50 \mathrm{~cm}$. There are two possible values for the length (in $\mathrm{cm}$ ) of the sides of the smallest square.
Which of the following are they?

A 2,32
B 4,42
C 4,34
D 32,40
E 34,42
Solution
A
We let $S, T, U$ and $V$ be the points shown in the figure.

We let the side length of the smallest square be $x \mathrm{~cm}$.

It follows that the middle square has sides of length $(x+8) \mathrm{cm}$
Since the lines $Q U$ and $R V$ are both perpendicular to $U V$ they are parallel lines.

Therefore, as $P Q R$ is a straight line, the corresponding angles $\angle P Q S$ and $\angle Q R T$ are equal.
Because the figure is made up of squares, $\angle P S U=\angle Q T V=90^{\circ}$. Therefore, as angles on a line add up to $180^{\circ}$, we have $\angle P S Q=\angle Q T R=90^{\circ}$. It follows that the right-angled triangles $P S Q$ and $Q T R$ are similar because they have the same angles. Therefore
$$
\frac{Q S}{P S}=\frac{R T}{Q T}
$$

Now $Q S$ has length $8 \mathrm{~cm}, P S$ has length $x \mathrm{~cm}, R T$ has length $(50-(x+8)) \mathrm{cm}=(42-x) \mathrm{cm}$ and $Q T$ has length $(x+8) \mathrm{cm}$. It follows that
$$
\frac{8}{x}=\frac{42-x}{x+8} .
$$

Multiplying both sides of this equation by $x(x+8)$, we see that this last equation is equivalent to
$$
8(x+8)=x(42-x),
$$
that is,
$$
8 x+64=42 x-x^{2} .
$$

This last equation may be rearranged to give
$$
x^{2}-34 x+64=0 \text {. }
$$

We can now factorize the quadratic in this equation to give
$$
(x-32)(x-2)=0 \text {. }
$$

Therefore
$$
x=2 \text { or } x=32 .
$$

We deduce that the possible lengths, in $\mathrm{cm}$, of the sides of the smallest square are 2 and 32 .