After an extended 1+ month break, we’re back with a vengeance. Here’s a link to the problems. No code this week.

**Riddler Express**

We know the width of the top rectangle since we already have its area and length: . From here, we can determine the width of the lower rectangle by adding the two width differences: . This implies that the height of the bottom rectangle is , giving us our final answer of **4 inches**^{1}.

**Riddler Classic**

Let the height and width of the unknown rectangle be and , respectively. Now, we can derive two separate equations from the top-left and bottom-right rectangles:

(1)

This is a system of equations with two variables and two unknowns which we can solve for using the quadratic equation (or, if you’re lazy like me, WolframAlpha):

(2)

The third rectangle will now allow us to determine which of the prior two is the correct solution. The first solution set (unknown area computes to ) gives us as the width of the upper-right rectangle, which is valid since . The second solution set (unknown area computes to ) gives us as the width, which is invalid since . Hence, the area of the unknown rectangle is **32 square inches**.

^{1} Shame on you, FiveThirtyEight, for not using the metric system!