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Problem of the Week
Problem A and Solution
Perimeter and Area

Problem

The diagram below shows the shape, and some of the dimensions, of an L-shaped parking lot. All angles in the diagram are right angles.

A shape composed of a larger rectangle with a smaller rectangle placed along the right side at the bottom. The top side of the larger rectangle is 40 m long. The left side of the larger rectangle is 20 m. The bottom sides of the two rectangles form a straight line with total length of 60 m. The right side of the smaller rectangle is 10 m.

  1. What is the perimeter of the parking lot?

  2. What is the area of the parking lot?


Solution

To determine the perimeter and area of the parking lot, we add two dashed lines that extend the sides labelled A and B below. This divides the L-shaped lot into three rectangles, labelled \(1\), \(2\), and \(3\), as shown below.

The top side of the smaller rectangle on the right is labelled B. A dashed line is drawn extending side B to the left which cuts the shape into a top piece and a bottom piece. The top piece is a rectangle labelled 1 and the right side of this rectangle is labelled A. A second dashed line is drawn extending side A down dividing the bottom piece in two. The bottom left piece is rectangle 2 and the bottom right piece is rectangle 3.

Since all angles are right angles, we know that the dashed line that is an extension of side B is \(40\) m in length, and the dashed line that is an extension of side A is \(10\) m in length.

We can now see that the length of side A is \(20 - 10 = 10\) m, and the length of side B is \(60 - 40 = 20\) m. A diagram with all of these dimensions included is shown below.

We now have enough information to calculate the perimeter and area of the parking lot.

  1. The perimeter of the parking lot is \(20 + 40 + 10 + 20 + 10 + 60 = 160\) m.

  2. The area of the parking lot is equal to the sum of the areas of the three rectangles, rectangles 1, 2, and 3.
    Rectangles 1 and 2 are each \(40~\text{m}\) by \(10~\text{m}\). Therefore, rectangles 1 and 2 each have an area of \(40 \times 10 = 400~\text{m}^2\).
    Rectangle 3 is \(20~\text{m}\) by \(10~\text{m}.\) Therefore, rectangle 3 has an area of \(20 \times 10 = 200~\text{m}^2\).
    Therefore, the total area of the parking lot is \(400 + 400 + 200 = 1000~\text{m}^2\).

    Alternatively, we could divide the diagram into two parts, a \(40~\text{m}\) by \(20~\text{m}\) rectangle and a \(20~\text{m}\) by \(10~\text{m}\) rectangle.

    A dashed line is drawn dividing the shape into two pieces: the larger rectangle on the left and the smaller rectangle on the right.

    The area of the \(40~\text{m}\) by \(20~\text{m}\) rectangle is \(40 \times 20 = 800~\text{m}^2\).
    The area of the \(20~\text{m}\) by \(10~\text{m}\) rectangle is \(20 \times 10 = 200~\text{m}^2\).
    Therefore, the total area of the parking lot is \(800 + 200 = 1000~\text{m}^2\).