How Do You Like Them Rectangles?

Riddler Express

In the staircase of three rectangles below, find the height marked with a question mark.

Solution

The three rectangles share a left edge, so their widths grow by the marked overhangs. The top rectangle has area $24$ and height $4$, hence width $24/4 = 6$. The middle rectangle juts out $3$ further and the bottom one juts out another $2$, so the bottom rectangle is $6 + 3 + 2 = 11$ wide. Its area is $44$, so the unknown height is $$\frac{44}{11} = \boxed{4 \text{ in.}}$$ The bottom rectangle is exactly as tall as the top one. (The middle area, $34$, is extra information, not needed for the height.)

The computation

Chain the widths down the staircase and divide the bottom area by the bottom width.

top_width = 24 / 4                 # top area / top height
bottom_width = top_width + 3 + 2   # add the two marked overhangs
print(44 / bottom_width)           # 4.0 in

Riddler Classic

Four rectangles are arranged inside a frame $14$ inches wide and $11$ inches tall, as below. Three of their areas are given; find the area of the shaded rectangle.

Solution

Let the shaded rectangle be $x$ wide and $y$ tall. Along the bottom it sits beside the $45$-rectangle to fill the $14$-inch width, both of height $y$, and up the left side it sits below the $32$-rectangle to fill the $11$-inch height, both of width $x$: $$(14 - x)\,y = 45, \qquad (11 - y)\,x = 32.$$ Subtracting the second from the first clears the $xy$ term, giving $14y - 11x = 13$. Substituting $y = (13 + 11x)/14$ into the first equation yields $$11x^2 - 141x + 448 = 0, \qquad x = \frac{141 \pm 13}{22} = 7 \ \text{ or } \ \tfrac{64}{11}.$$ The two roots describe two candidate figures, and the top-right $34$-rectangle settles which is real: with height $11 - y$ it has width $34/(11-y)$, and starting at $x$ it must fit inside the $14$-inch frame, $x + 34/(11-y) \le 14$. The root $x = 7,\ y = \tfrac{45}{7}$ needs width $14.44$ and overflows; the root $x = \tfrac{64}{11},\ y = \tfrac{11}{2}$ needs only $12$ and fits. So the shaded area is $$xy = \frac{64}{11}\cdot\frac{11}{2} = \boxed{32 \text{ in.}^2}.$$

The computation

Solve the two area equations symbolically and keep the root whose top-right rectangle fits inside the $14$-inch width.

import sympy as sp
x, y = sp.symbols('x y', positive=True)
sols = sp.solve([sp.Eq((14 - x) * y, 45), sp.Eq((11 - y) * x, 32)], [x, y])
for xv, yv in sols:
    fits = xv + 34 / (11 - yv) <= 14            # top-right rectangle fits?
    print(xv, yv, "area", xv * yv, "fits" if fits else "overflows")