Question Video: Finding the Area of a Rectangle Using the Distance between Two Points Formula

Video Transcript

Given that 𝐴𝐵𝐶𝑂 is a rectangle, find the length of line segment 𝐴𝐶 and the area of the rectangle.

In this question, we can see this rectangle 𝐴𝐵𝐶𝑂 and we need to find a length and the area. We might notice that we’re not given any lengths in this diagram. So in order to answer this question, we’ll need to look a little bit more closely at the vertices.

Firstly, this vertex of 𝑂 must be the coordinate or ordered pair of zero, zero. This other important piece of information that 𝐵 is at 12, five will help us work out the coordinates of vertices 𝐴 and 𝐶. We know that the line 𝑂𝐴 lies on the 𝑥-axis, which is a horizontal line. And so the line 𝐵𝐶 must also be a horizontal line.

We must remember that the definition of a rectangle is that all interior angles are 90 degrees. So we’ll have right angles at angles 𝑂, 𝐴, 𝐵, and 𝐶. And we can be sure that 𝑂𝐶 and 𝐴𝐵 will be vertical lines. So let’s look at this vertex 𝐴 in relation to 𝐵. We know that 𝐵 lies at 12 on the 𝑥-axis, and therefore 𝐴 must also lie at 12 on the 𝑥-axis. We can see that point 𝐴 sits on the 𝑥-axis, so the 𝑦-value will be zero. We can alternatively think of this that the line segment 𝑂𝐴 must be 12 units long.

Next, let’s think about this vertex 𝐶. It lies on the 𝑦-axis. Therefore, the 𝑥-coordinate will be zero, and the 𝑦-value will be the same as that of 𝐵. So the coordinate of 𝐵 must be zero, five. Now that we’ve found the coordinate values of all four vertices, let’s see if we can find the length of this line segment 𝐴𝐶. This is the diagonal of the rectangle going from 𝐴 to 𝐶.

As we know the coordinate values for vertices 𝐴 and 𝐶, we can recall the formula for the distance between two points. This tells us that, for any pair of coordinates, 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance can be found by the square root of 𝑥 two minus 𝑥 one all squared plus 𝑦 two minus 𝑦 one all squared. So let’s take our two vertices. We can make 𝐴 our 𝑥 one, 𝑦 one values and 𝐶 our 𝑥 two, 𝑦 two values. But it doesn’t matter which way round we write these just so long as we take consistently from each coordinate.

So plugging these into the formula would give us 𝑑 equals the square root of zero minus 12 squared plus five minus zero squared. Simplifying this, we have the square root of negative 12 squared plus five squared. Negative 12 squared will give us 144, and five squared gives us 25. So adding these together would give us 169. Finding the square root of 169 would give us a value of 13. Even though we’ve done some squaring, we’re not finding an area, so the units here will be length units. Therefore, we’ve found our first answer that the length of line segment 𝐴𝐶 would be 13 length units.

Next, let’s see how we can find the area of this rectangle. We should remember that to find the area of a rectangle, we multiply the length by the width. Earlier in the question, we established that 𝑂𝐴 must be 12 units long as 𝐴 lies at 12 on the 𝑥-axis and 𝑂 lies at zero on the 𝑥-axis. In a similar way, we can look at the length of the line segment 𝐴𝐵 and observe that it goes from zero on the 𝑦-axis to five on the 𝑦-axis, which means that the line segment 𝐴𝐵 must be five units long.

Therefore, in order to find the area, we would take our length and width of 12 and five and multiply them together, which gives us a value of 60. And this time, as we’re working with an area, then our units will be area units. So here we gave our two answers. The line segment 𝐴𝐶 was 13 length units, and the area of the rectangle is 60 area units.