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Problem of the Week
Problem C and Solution
A Rectangle of Rectangles

Problem

Large rectangle \(JKLM\) is formed by twelve identical smaller rectangles, as shown.

5 smaller rectangles are stacked on top of each other. Their combined widths form side JM of the large rectangle. A sixth smaller rectangle is to the right of the stack and arranged so that its length is adjacent to the widths of the 5 stacked rectangles. To its right, this arrangement of 6 rectangles is repeated.

If the area of \(JKLM\) is \(540\text{ cm}^2\), then determine the dimensions of the smaller rectangles.

Solution

Let \(x\) be the width of one of the smaller identical rectangles, in cm. Five of the smaller rectangles are stacked on top of each other forming \(JM\), so \(JM=x+x+x+x+x=5x\). Since \(JKLM\) is a rectangle, \(JM=KL=5x\). Thus, \(5x\) is also the length of a smaller rectangle. Therefore, a smaller rectangle is \(5x\text{ cm}\) by \(x\text{ cm}\).

From here, we proceed with two different solutions.

Solution 1

Since \(JKLM\) is formed by twelve identical smaller rectangles, the area of rectangle \(JKLM\) is equal to \(12\) times the area of one of the smaller rectangles. \[\begin{aligned} \text{Area }JKLM &= 12 \times \text{Area of one smaller rectangle}\\ 540&=12 \times 5x \times x\\ 540&=60 \times x^2 \end{aligned}\] Dividing both sides by \(60\), we obtain \(x^2=9\). Since \(x\) is the width of a smaller rectangle, \(x>0\), and so \(x=3\) follows.

Thus, the width of a smaller rectangle is \(x=3\text{ cm}\) and the length of a smaller rectangle is \(5x=5(3)=15\text{ cm}\).

Therefore, the smaller rectangles are each \(15\text{ cm}\) by \(3\text{ cm}\).

Solution 2

Side length \(ML\) is made up of the lengths of two of the smaller rectangles plus the widths of two of the smaller rectangles. Therefore, \(LM=5x+5x+x+x=12x\) and rectangle \(JKLM\) is \(12x\text{ cm}\) by \(5x\text{ cm}\).

To find the area of \(JKLM\) we multiply the length \(ML\) by the width \(JM\). \[\begin{aligned} \text{Area }JKLM &=ML \times JM\\ 540&= (12x) \times (5x)\\ 540&=12 \times 5 \times x \times x\\ 540&=60 \times x^2 \end{aligned}\] Dividing both sides by \(60\), we obtain \(x^2=9\). Since \(x\) is the width of a smaller rectangle, \(x>0\), and so \(x=3\) follows.

Thus, the width of a smaller rectangle is \(x=3\text{ cm}\) and the length of a smaller rectangle is \(5x=5(3)=15\text{ cm}\).

Therefore, the smaller rectangles are each \(15\text{ cm}\) by \(3\text{ cm}\).