Calculating the area of a complex shape

Jenny loves cars, and works at a racetrack. The track is old, and needs to be resurfaced with asphalt. Jenny will calculate how much asphalt is needed. Here is a plan of the.. kind of ‘oval’ track.

The perimeter around the inside of the track is 400 metres. There are two straight stretches of 100 metres each. And the width is 16 metres. There is no exact geometric shape we could use for the entire racetrack. It consists of several different shapes, two rectangles, and two arcs.

So the track is a complex shape. And we have to calculate the straight stretches and the curves separately. The straight stretches look like rectangles. 100 metres long and 16 metres wide. The area of a straight stretch is therefore: 100 metres times 16 metres, which is: 1600 square metres.

There are two similar straight stretches, so together they are 1600 times 2, that is 3200 square metres. But these curves, these arcs? How do we calculate the area of them? If we put both arcs together like this, we get two circles, one inside the other. And the total area of the two curved stretches of track is the outer circle’s area minus the inner circle’s area.

The area of a circle is pi times the radius squared. So we need to find out the radius. It’s tricky, but it is doable. The inner perimeter of the whole racing track is 400 meters. If we subtract the straight stretches, each 100 meters, there are 200 meters of perimeter remaining.

That distance is the perimeter of the inner circle. The perimeter of the circle, 200 metres, is pi times the diameter. Divide by pi on both sides of the equals sign. The diameter is 200 divided by 3,14 [3.14], which is approximately 64 metres. And the radius is half that diameter.

Half of 64 is 32. The inner circle has the area: pi times 32 squared, which is approximately 3217 square metres. The radius of the outer circle is 32 plus the track’s width, 16 meters, which is 48 metres. The outer circle then has the area pi times 48 squared, which is approximately 7238 square metres. The two curves of the racing track have the area of the outer circle minus the inner circle.

Jenny calculates that: 4021 square metres. So, how much asphalt does she need? Add the area of the straight stretches, 3200 plus the curves, 4021. That equals 7221 square metres. So 7221 square metres will be resurfaced!

Here comes Jenny’s boss. Ah! The cars need a place where they go in and refuel and change tires, a pit. That has the shape of a trapezoid. The short sides of the trapezoid are the same length, so it is isosceles.

The sides are 13 metres. On the racing track side it’s 100 metres, and the other parallel side is 90 metres. The width of the pit is 12 metres. We get the area by adding the two long sides, and then dividing by two. 100 plus 90 divided by 2 which is 95.

Then we multiply the answer by the width of the track, 12. The area of the trapezoid is 1140 square metres. Together with the rest of the racing track, there are 7221 plus 1140 about 8400 square metres to be resurfaced. That was a lot of calculation for Jenny to find the area of the whole racetrack, the complex shape - plus the pit! Now she deserves a break.

But wait, it hasn’t been... ...resurfaced yet!