Problem of the Week
Problem
B and Solution
Puzzling
Areas
Problem
Four rectangles meet at a common vertex, as shown in the diagram.
Two top rectangles are
placed side by side with their bottom sides along a horizontal line. Two
bottom rectangles are placed side by side with their top sides along the
same horizontal line, arranged so the four rectangles share a
vertex.
The four rectangles have different sizes and shapes. Some
measurements of the rectangles are given and others are represented by
the variables \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\), as follows:
The top right rectangle has area \(576
\text{ cm}^{2}\). The length of its top side is \(c \text{ cm}\) and the length of its right
side is \(b \text{ cm}\).
The bottom left rectangle has area \(f
\text{ cm}^{2}\). The length of its left side is \(e \text{ cm}\) and the length of its bottom
side is \(24 \text{ cm}\).
The top left rectangle has area \(270
\text{ cm}^{2}\). The length of its top side is \(a \text{ cm}\). Its bottom side is \(6 \text{ cm}\) shorter than the top side of
the bottom left rectangle, and its right side is \(9 \text{ cm}\) shorter than the left side
of the top right rectangle.
The bottom right rectangle has area \(216 \text{ cm}^{2}\). The length of its top
side is half the length of the bottom side of the top right rectangle.
The length of its right side is \(d \text{
cm}\). Its left side is \(12 \text{
cm}\) shorter than the right side of the bottom left
rectangle.
Determine the values of \(a\), \(b\), \(c\), \(d\), \(e\), and \(f\).
Solution
We will call the top left rectangle \(A\), the top right rectangle \(B\), the bottom right rectangle \(C\), and the bottom left rectangle \(D\).
Since the length of the bottom side of \(D\) is \(24\text{
cm}\), the length of the top side of \(D\) is also \(24\text{ cm}\). Hence, the length of the
bottom side of \(A\) is \(24-6=18\text{ cm}\). Therefore, the length
of the top side of \(A\) is \(18\text{ cm}\) and so \(a=18\).
Since the area of \(A\) is \(270\text{ cm}^2\) and the length of the
bottom side of \(A\) is \(18\text{ cm}\), then the length of the
right side of \(A\) is \(270\div 18=15\text{ cm}\). Therefore, \(b= 9 + 15 = 24\).
Since the area of \(B\) is \(576\text{ cm}^2\) and \(b=24\), then \(c=576\div 24=24\).
Since the length of the top side of \(B\) is \(24\text{
cm}\), then the length of the bottom side is also \(24\text{ cm}\). Thus, the length of the top
side of \(C\) is half of \(24\text{ cm}\) or \(12\text{ cm}\). Since the area of \(C\) is \(216\text{ cm}^2\), then \(d=216\div 12=18\).
Thus, \(e = d + 12 = 18 + 12 =
30\).
Finally, the area of \(D\) is equal
to \(24\times e = 24(30) = 720\text{
cm}^2\) and so \(f=720\).
Note: this puzzle is adapted from "Area Mazes" by Naoki
Inaba.
