## Riddler Express

Area puzzle ## Solution

The length of the top rectangle = $24/4 = 6 \ in$.

The length of the bottom rectangle = $6+3+2=11 \ in$.

The height of the bottom rectangle = $44/11 = 4 \ in$

## Riddler Classic

Area puzzle ## Solution

Let $x in$ and $y in$ be the length and height of the shaded rectangle.

We have the following constraints:

$$\begin{eqnarray} (14-x)y &=& 45 \ (11-y)x &=& 32 \ \frac{66}{11-y} &<& 14 \implies y < \frac{44}{7} \end{eqnarray}$$

Subtracting (2) from (1) we have $14y - 11x = 13$.

Substituting $y = \frac{13 + 11x}{14}$ in (1), we have

$$\begin{eqnarray*} 13 + 11x - 13x/14 - 11x^2/14 &=& 45 \ \implies 11x^2 - 141x + 448 &=& 0 \ \implies x &=& \frac{141 \pm \sqrt{141^2 - 4 \cdot 11 \cdot 448}}{2 \cdot 11} \ \implies x = 7 \ and \ y = \frac{45}{7} \ &or& \ x = \frac{64}{11}\ and \ y = \frac{11}{2} \end{eqnarray*}$$

The first solution has to be discarded because of constraint (3).

The area of the shaded region is $xy = \frac{64}{11} \frac{11}{2} = 32 \ in.^2$