Video Transcript
π΄π΅πΆπ· is a rectangle in which π΄π· equals 21 centimeters, π΄π equals nine centimeters, and ππ equals 12 centimeters. Calculate the perimeter of triangle πππΆ.
We can start by filling in any length information that weβre given onto the diagram. π΄π· is 21 centimeters, π΄π is nine centimeters, and ππ is 12 centimeters. Weβre now asked to find the perimeter of triangle πππΆ, 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 donβt know any of these three lengths on this triangle πππΆ. So letβs see if we can find some way to figure those out. Letβs begin by seeing if we can work out this length of πΆπ. And we should notice that weβre told that π΄π΅πΆπ· is a rectangle.
In a rectangle, we have two pairs of opposite sides parallel. So the length of π΄π΅ is parallel to this length π·πΆ. The bases of π΄π· and πΆπ΅ will also be parallel. Because we were given on the marking of the diagram that the line segment π΄π΅ is parallel to the line segment ππ, 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 π and π will all be 90 degrees as well.
Weβll 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 πΆπ, it will in fact be equal to the length of the line segment π·π. And we can work out π·π. Weβre told that π·π΄ is 21 centimeters and π΄π is nine centimeters. So subtracting those would give us 12. So π·π is 12 centimeters, and so is πΆπ.
Now weβve found one length on this triangle πππΆ, weβll need to work out another way to help us find the other two sides. Letβs see if we can use this other triangle, triangle πππ΄, to help us work out some of the missing lengths. We can see that these two triangles are different sizes. So letβs check if theyβre similar.
Remember that when two shapes are similar, that means that corresponding sides are in the same proportion and corresponding angles are congruent. Letβs look at some of the angles. If we take this angle π΄ππ and this angle πΆππ, we actually know that theyβre equal because theyβre opposite angles. Next, we can say that angle π΄ππ is equal to angle πΆππ 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 ππ΄π is equal to angle ππΆπ. Thatβs because weβve got our two parallel lines π΄π· and πΆπ΅ and the transversal π΄πΆ. 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 πππΆ. We notice that we have a pair of corresponding sides. They are πΆπ and π΄π. 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 ππ, then we find the corresponding side in the smaller triangle. And it will be ππ. We can therefore write that πΆπ over π΄π is equal to ππ over ππ.
We couldβve 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 ππ over 12.
Before we take the cross product, we can simplify. So weβll get four times 12 is equal to three times ππ. We can then simplify and divide both sides by three. And we find that the length of ππ is 16 centimeters. In order to find the perimeter, however, weβll need to find the length of this final side πΆπ. Unfortunately, however, weβre not given the corresponding length in triangle πππ΄. Thatβs the side π΄π. So weβll 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 πΆπ 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 letβs say that we define our length πΆπ to be π₯ centimeters. We could then write that π₯ 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 π₯, 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 πΆπ on this triangle is 20 centimeters.
Alternatively, we couldβve observed that the sides πΆπ and ππ form the three and four part of a three-four-five Pythagorean triple. Of course, as triangle ππ΄π is similar, we should also see this three-four-five Pythagorean triple, giving us that the length of π΄π, if we wish to work it out, is 15 centimeters.
Finally, after all those calculations, weβre ready to work out the perimeter of triangle πππΆ. That means weβre going to add the three lengths, 20 centimeters, 16 centimeters, and 12 centimeters, giving us a final answer of 48 centimeters.