# Question Video: Finding the Length of the Diagonal of a Trapezoid given Its Sides’ Lengths Mathematics

In trapezoid 𝐴𝐵𝐶𝐷, sides line segment 𝐴𝐷 and line segment 𝐵𝐶 are parallel, and its diagonals intersect at 𝑀. Given that 𝐴𝐷 = 66, 𝐵𝐶 = 33, and 𝐴𝐶 = 75, what is the length of line segment 𝑀𝐴?

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### Video Transcript

In trapezoid 𝐴𝐵𝐶𝐷, sides line segment 𝐴𝐷 and line segment 𝐵𝐶 are parallel, and its diagonals intersect at 𝑀. Given that 𝐴𝐷 equals 66, 𝐵𝐶 equals 33, and 𝐴𝐶 equals 75, what is the length of line segment 𝑀𝐴?

Let’s begin this question by having a look at the diagram. We have a trapezoid 𝐴𝐵𝐶𝐷, and we’re given that there is a pair of parallel sides 𝐴𝐷 and 𝐵𝐶. We’re also given some lengths, which we can fill onto the diagram. 𝐴𝐷 is 66, 𝐵𝐶 is 33, and 𝐴𝐶 is 75. We’re asked to work out the length of line segment 𝑀𝐴, which is part of the diagonal 𝐴𝐶. At this point, we haven’t got enough information to work out the length of line segment 𝑀𝐴. So let’s consider if we have any similar triangles in this trapezoid. Specifically, let’s look at this triangle 𝑀𝐵𝐶 and this triangle 𝑀𝐷𝐴. Let’s check if triangle 𝑀𝐵𝐶 is similar to triangle 𝑀𝐷𝐴. But first, let’s recall the definition for similar triangles.

Similar triangles have corresponding pairs of angles congruent and corresponding pairs of sides in proportion. One way that we could prove that triangles are similar is by using the SSS rule, which checks if three pairs of corresponding sides have the same proportion. However, we’re not given enough information about the sides, so let’s see if we can apply the AA similarity criterion. For this, we would need to check that there are two pairs of corresponding angles congruent. Let’s begin with the angle 𝑀𝐵𝐶. Because we have a pair of parallel lines 𝐵𝐶 and 𝐴𝐷, then we actually have an angle that’s equal to angle 𝑀𝐵𝐶. It’s this angle at 𝑀𝐷𝐴. Because of the parallel lines and the transversal 𝐵𝐷, then these two angles are alternate.

In the same way, if we use our two parallel lines and the transversal this time of 𝐴𝐶, we can say that angle 𝑀𝐶𝐵 must be equal to angle 𝑀𝐴𝐷 as these angles are alternate. Since we’ve found two pairs of corresponding angles congruent, then this fulfills the AA criterion and proves that triangle 𝑀𝐵𝐶 is similar to triangle 𝑀𝐷𝐴. Of course, we could’ve also shown that angle 𝐵𝑀𝐶 is equal to angle 𝐷𝑀𝐴 as we have a pair of opposite angles. Any two of these three angle pairs would demonstrate similarity.

Now that we know that these two triangles are similar, let’s see how we can use this to help us work out this length of the line segment 𝑀𝐴. Let’s look at these corresponding sides, 𝐵𝐶 and 𝐴𝐷. Because these two lines are in proportion, then that means we could actually work out the scale factor of the smaller triangle to the larger triangle. 66 is double 33, so that means that the scale factor from the smaller triangle 𝑀𝐵𝐶 to the larger triangle 𝑀𝐷𝐴 must be two. And so the length of 𝑀𝐴 that we wish to find out must be double the length of the corresponding side, which would be side 𝐶𝑀.

But we don’t actually know the length of 𝐶𝑀. So let’s see if we can use the fact that the whole length of 𝐴𝐶 is 75. We have worked out that the scale factor from triangle 𝑀𝐵𝐶 to triangle 𝑀𝐷𝐴 is two. So that means that we could write the ratio of the line 𝐶𝑀 to the line 𝐴𝑀 as the ratio of one to two. In order to divide the length 𝐴𝐶 of 75 in the ratio one to two, we would start by dividing 75 by three, which would give us 25.

Therefore, we know that the one part of 𝐶𝑀 must be equal to 25 and the two part of 𝐴𝑀 must be equal to two times 25 which is 50. We can give the answer then that the length of line segment 𝑀𝐴 is 50, and if we needed to give a unit, these would be length units.