Video Transcript
𝐴𝐵𝐶𝐷 is a rectangle in which 𝐴𝐷 equals 21 centimeters, 𝐴𝑋 equals nine centimeters, and 𝑋𝑀 equals 12 centimeters. Calculate the perimeter of triangle 𝑌𝑀𝐶.
We can start by filling in any length information that we’re given onto the diagram. 𝐴𝐷 is 21 centimeters, 𝐴𝑋 is nine centimeters, and 𝑋𝑀 is 12 centimeters. We’re now asked to find the perimeter of triangle 𝑌𝑀𝐶, which is this triangle at the base of this rectangle.
We should remember that the perimeter is the distance around the outside edge or the total of all the outside edge lengths. At this point, however, we don’t know any of these three lengths on this triangle 𝑌𝑀𝐶. So let’s see if we can find some way to figure those out. Let’s begin by seeing if we can work out this length of 𝐶𝑌. And we should notice that we’re told that 𝐴𝐵𝐶𝐷 is a rectangle.
In a rectangle, we have two pairs of opposite sides parallel. So the length of 𝐴𝐵 is parallel to this length 𝐷𝐶. The bases of 𝐴𝐷 and 𝐶𝐵 will also be parallel. Because we were given on the marking of the diagram that the line segment 𝐴𝐵 is parallel to the line segment 𝑋𝑌, then actually what we have here is all three of these vertical lines are parallel. And since we have a rectangle, that means that the angles at the vertices in the rectangle are 90 degrees. And so the angles created at 𝑋 and 𝑌 will all be 90 degrees as well.
We’ll use this information about these 90-degree angles a little more in this question as we go through it. But notice that if we wanted to find this length of 𝐶𝑌, it will in fact be equal to the length of the line segment 𝐷𝑋. And we can work out 𝐷𝑋. We’re told that 𝐷𝐴 is 21 centimeters and 𝐴𝑋 is nine centimeters. So subtracting those would give us 12. So 𝐷𝑋 is 12 centimeters, and so is 𝐶𝑌.
Now we’ve found one length on this triangle 𝑌𝑀𝐶, we’ll need to work out another way to help us find the other two sides. Let’s see if we can use this other triangle, triangle 𝑋𝑀𝐴, to help us work out some of the missing lengths. We can see that these two triangles are different sizes. So let’s check if they’re similar.
Remember that when two shapes are similar, that means that corresponding sides are in the same proportion and corresponding angles are congruent. Let’s look at some of the angles. If we take this angle 𝐴𝑀𝑋 and this angle 𝐶𝑀𝑌, we actually know that they’re equal because they’re opposite angles. Next, we can say that angle 𝐴𝑋𝑀 is equal to angle 𝐶𝑌𝑀 as we showed earlier that these will both be 90-degree angles. Having these two pairs of corresponding angles congruent is enough to prove the AA similarity criterion. This means that, yes,s these two triangles are similar.
We could have also demonstrated that the third pair of angles are congruent. Angle 𝑋𝐴𝑀 is equal to angle 𝑌𝐶𝑀. That’s because we’ve got our two parallel lines 𝐴𝐷 and 𝐶𝐵 and the transversal 𝐴𝐶. So showing any of these two pairs of angles are congruent does demonstrate that the two triangles are similar.
We can now use this fact about the similar triangles to help us work out some of the missing lengths in triangle 𝑌𝑀𝐶. We notice that we have a pair of corresponding sides. They are 𝐶𝑌 and 𝐴𝑋. And because these triangles are similar, we know that these sides are in a proportion. And that proportion will be equal to the proportion of any of the two other pairs of corresponding sides. For example, if we want to find the length of the side 𝑀𝑌, then we find the corresponding side in the smaller triangle. And it will be 𝑀𝑋. We can therefore write that 𝐶𝑌 over 𝐴𝑋 is equal to 𝑀𝑌 over 𝑀𝑋.
We could’ve written these ratios or fractions with the numerators and denominators flipped. But in that case, we need to make sure that we essentially keep the same triangle lengths on either always on the numerator or always on the denominator. We can then fill in the values that we were given for the lengths. So we have 12 over nine is equal to the length 𝑀𝑌 over 12.
Before we take the cross product, we can simplify. So we’ll get four times 12 is equal to three times 𝑀𝑌. We can then simplify and divide both sides by three. And we find that the length of 𝑀𝑌 is 16 centimeters. In order to find the perimeter, however, we’ll need to find the length of this final side 𝐶𝑀. Unfortunately, however, we’re not given the corresponding length in triangle 𝑋𝑀𝐴. That’s the side 𝐴𝑀. So we’ll need to find a different method.
As we have a right triangle, we could apply the Pythagorean theorem. We could do it in one of two ways. We could find this length in the smaller triangle and then work out its corresponding side on the larger triangle. However, perhaps the quickest way is simply to find the length of 𝐶𝑀 straight away on the larger triangle by applying the Pythagorean theorem.
The Pythagorean theorem tells us that, for every right triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides. So let’s say that we define our length 𝐶𝑀 to be 𝑥 centimeters. We could then write that 𝑥 squared is equal to 16 squared plus 12 squared. 16 squared is 256. 12 squared is 144. And adding those gives us 400. To find 𝑥, we need to perform the inverse operation, which is taking the square root. And the square root of 400 is 20. And so that means that the final length of 𝐶𝑀 on this triangle is 20 centimeters.
Alternatively, we could’ve observed that the sides 𝐶𝑌 and 𝑀𝑌 form the three and four part of a three-four-five Pythagorean triple. Of course, as triangle 𝑋𝐴𝑀 is similar, we should also see this three-four-five Pythagorean triple, giving us that the length of 𝐴𝑀, if we wish to work it out, is 15 centimeters.
Finally, after all those calculations, we’re ready to work out the perimeter of triangle 𝑌𝑀𝐶. That means we’re going to add the three lengths, 20 centimeters, 16 centimeters, and 12 centimeters, giving us a final answer of 48 centimeters.
