Question Video: Finding the Perimeter and Area of a Parallelogram on a Coordinate Grid

Video Transcript

A parallelogram has vertices at the points 𝐴, 𝐵, 𝐶, and 𝐷 with coordinates negative one, one; one, three; three, negative one; and one, negative three, respectively. Work out the perimeter of the parallelogram 𝐴𝐵𝐶𝐷. Give your solution to one decimal place. By drawing a rectangle through the vertices of the parallelogram, or otherwise, work out the area of the parallelogram 𝐴𝐵𝐶𝐷.

In this question, we’re given four coordinates for four vertices of a parallelogram. As these are relatively simple coordinates, perhaps the most sensible thing to do is to plot these. So, let’s take a grid and plot these four points. So, here, we have 𝐴 at negative one, one; 𝐵 at one, three; 𝐶 at three, negative one; and 𝐷 at one, negative three. We can then join these four vertices to create the parallelogram 𝐴𝐵𝐶𝐷. We can observe too that this is a parallelogram, as that’s a quadrilateral that has opposite sides parallel. It’s also a good check that we’ve plotted these points correctly.

Let’s now look at the first part of this question which asks us to work out the perimeter of the parallelogram. We can recall that the perimeter of a shape is given by the distance around the outside edge. Unfortunately, with this parallelogram, this isn’t as simple as if the sides were given as horizontal or vertical lines. However, we still can work out the perimeter of this parallelogram by finding the lengths of the sides. Let’s take this length 𝐴𝐵. We could work this out using the Pythagorean theorem or we could use a formula that applies the Pythagorean theorem.

This formula allows us to find the distance between two coordinates 𝑥 sub one, 𝑦 sub one and 𝑥 sub two, 𝑦 sub two. The distance is given as the square root of 𝑥 sub two minus 𝑥 sub one squared plus 𝑦 sub two minus 𝑦 sub one squared. We can take the coordinate of 𝐴 to be our 𝑥 sub one, 𝑦 sub one values and the coordinate of 𝐵 to be the 𝑥 sub two, 𝑦 sub two values. We then plug these values into the formula to find the distance 𝐴𝐵.

So, 𝐴𝐵 is equal to the square root of one subtract negative one squared plus three subtract one squared. Simplifying this, if we look at the first set of parentheses, one subtract negative one is equivalent to one plus one, which is two. And in the second set of parentheses, three subtract one is also two. So, we need to take the square root of two squared plus two squared. Then, as two squared is equal to four, we have four plus four, which is eight. So, 𝐴𝐵 is equal to the square root of eight.

Although the final answer for this part of the question is needed to one decimal place, we’ll keep this in the square root form as the square root of eight. As we know that this is a parallelogram, then this means that opposite sides are congruent. So, the length of 𝐶𝐷 is also equal to the square root of eight. We can then use this same formula to work out the length of 𝐴𝐷. It doesn’t matter which coordinate we designate with the 𝑥 sub one, 𝑦 sub one values. So, let’s make 𝐷 have the 𝑥 sub two, 𝑦 sub two values this time. Plugging these values into the formula to find the length of 𝐴𝐷, we have 𝐴𝐷 equals the square root of one subtract negative one squared plus negative three subtract one squared.

Simplifying this, we have 𝐴𝐷 is equal to the square root of two squared plus negative four squared. We know that two squared is four and negative four squared is 16. So, adding these together gives us 20, and finding the square root gives us root 20. So, we’ve now found that the length of 𝐴𝐷 and likewise the length of 𝐵𝐶 could be given as root 20. Notice that these lengths would all be given with length units.

Finally, to answer this question, we need to work out the perimeter. We know that there are two sides that have a length of root eight units and two sides that have a length of root 20 length units. As we need to give the answer to one decimal place, we can use our calculator to give us a value of 14.6011 and so on. And we then need to check our second decimal digit to see if it’s five or more, which means that our answer runs down to 14.6. Therefore, 𝐴𝐵𝐶𝐷 has a perimeter of 14.6. We weren’t given any units in the question, but if we needed to give them, these would be length units.

We’ll now clear some space so we can answer the second part of this question. In this question, we’re asked to find the area of the parallelogram 𝐴𝐵𝐶𝐷. You might recall that to find the area of a parallelogram, we work out the base multiplied by the perpendicular height. Let’s say that we took the base to be the length of 𝐴𝐷. To use this formula, we’d then need to work out the perpendicular length from 𝐶 to 𝐴𝐷. As this looks like it’s going to be rather complicated, let’s have a look again at this part of the question.

We’re actually given the hint to draw a rectangle through the vertices of the parallelogram. So, let’s see what that looks like. Here we have a rectangle. We know that it’s got four right angles, and we also know that opposite sides are parallel and congruent. But how does this help us to find the area of the parallelogram? Let’s go one step further and divide this rectangle and the parallelogram within it into two equal-sized pieces.

If we shade the left side of this parallelogram, we’ve actually created another shape, a triangle. And we should recall that the area of a triangle is equal to half multiplied by the base multiplied by the perpendicular height. To work out the area of this triangle on the left, triangle 𝐴𝐵𝐷, the base would be given by 𝐵𝐷. Counting the squares on the grid, the length of 𝐵𝐷 could be given as six units long. The perpendicular height would be two units long. So, the area of this triangle is half multiplied by six multiplied by two. This evaluates to six square units.

So, we’ve found the area of part of the parallelogram. But what about the other side, the triangle 𝐶𝐷𝐵? Well, we could go ahead and calculate the area of this triangle using the formula and noticing that this has the same perpendicular height of two units. But you may also have noticed that triangle 𝐴𝐵𝐷 and triangle 𝐶𝐷𝐵 are in fact congruent triangles. But either way, we would’ve found that both of these triangles have an area of six square units each. Therefore, to find the area of the whole parallelogram 𝐴𝐵𝐶𝐷, we take the two areas that it’s composed of, the two triangles with an area of six each, and add them together, which gives us a final answer of 12. And if we needed to give the units, they would be square units.