# Problem of the Week Problem B and Solution Mystery Dimensions

## Problem

Eight congruent rectangles are arranged to form a larger rectangle as shown.

1. If the congruent rectangles each have a length of $$6$$ cm and a width of $$3$$ cm, what is the perimeter of the larger rectangle?

2. Suppose that the congruent rectangles each have a longer side of length $$L$$ cm and a shorter side of length $$4$$ cm. Suppose also that the perimeter of the larger rectangle is $$64$$ cm.

1. What is the value of $$L$$?

2. What is the area of one of the eight congruent rectangles?

Extension: Can you solve part (b) without knowing that the length of the shorter side of each rectangle is $$4$$ cm? If so, how?

## Solution

1. Since each rectangle has a length of $$6$$ cm and a width of $$3$$ cm, the larger rectangle must have sides of lengths $$6+6=12$$ cm and $$3+6+3=12$$ cm. Thus, the perimeter of the larger rectangle is $$12+12+12+12 = 48$$ cm.

1. Since each rectangle has a longer side of length $$L$$ cm and shorter side of length $$4$$ cm, we can label our diagram to find the dimensions of the larger rectangle.

Using this, we determine that the lengths of the sides of the larger rectangle are $$L+L=2L$$ and $$4+L+4=L+8$$. Since we know the perimeter of the larger rectangle is $$64$$ cm, we can write the following equation.

\begin{align} 2L + 2L+(L+8) + (L+8) &=64\\ 6L+16&=64\\ 6L&=64-16\\ 6L&=48 \end{align} Since $$6\times 8=48$$, it follows that $$L=8$$ cm.

2. The area of a rectangle is equal to its length times its width. Thus, the area of each congruent rectangle is $$8\times 4 = 32~\text{cm}^2$$.

Extension Solution:

If we ignore the two rectangles on the top and the two rectangles on the bottom, we can see that two rectangles placed on top of each other horizontally have a height of $$L$$. Therefore, the shorter side of each rectangle equals half its longer side, or $$\frac{L}{2}$$. We can label our diagram to find the dimensions of the larger rectangle.

Using this, we determine that the larger rectangle has sides of length $$L+L=2L$$ and $$\frac{L}{2}+L+\frac{L}{2}=2L$$. So the larger rectangle is actually a square with side length $$2L$$. Since we know its perimeter is $$64$$ cm, it follows that $$2L+2L+2L+2L = 64$$, or $$8L = 64$$. Since $$8\times 8=64$$, it follows that $$L=8$$ cm. So, we can solve this problem without knowing the width of each rectangle.