Lesson Video: Right Triangle Altitude Theorem | Nagwa Lesson Video: Right Triangle Altitude Theorem | Nagwa

Lesson Video: Right Triangle Altitude Theorem Mathematics

In this video, we will learn how to use the right triangle altitude theorem, also known as the Euclidean theorem, to find a missing length.

15:32

Video Transcript

In this video, we’ll learn how to use the right triangle altitude theorem, also known as the Euclidean theorem, to find a missing length. This theorem is a useful tool to rewrite expressions involving the lengths of sides in a right triangle with a projection from the right angle onto the hypotenuse. In particular, it allows us to determine the lengths of sides in a right triangle given two of the lengths. Our plan of action in this video is we begin with a right triangle, adding squares to each side of the triangle, and use projections and the properties of congruent triangles to derive the Euclidean theorem. We’ll then use this theorem, together with the Pythagorean theorem, to derive a corollary. We’ll then see how the theorem and the corollary can be applied in some examples.

To derive the Euclidean theorem, we begin with a right triangle 𝐴𝐡𝐢 with right angle at 𝐴. We then project 𝐴 onto the side 𝐢𝐡 and call this point 𝐷 as shown. The line segment 𝐴𝐷 is perpendicular to side 𝐢𝐡. We now add to the diagram three squares given by each side of triangle 𝐴𝐡𝐢 and label the vertices of these squares as shown. We next want to continue our projection from 𝐴 to the side 𝐢𝐡 down to the side 𝐺𝐻. And we then add the line 𝐢𝐹 and 𝐴𝐺 to our diagram as shown. And we want to show now that triangle 𝐢𝐡𝐹 and triangle 𝐺𝐡𝐴 are congruent. We can do this by first noting that the measure of angle 𝐢𝐡𝐹 is equal to the measure of angle 𝐴𝐡𝐺. And this is true because these are both right angles added to angle 𝐴𝐡𝐢.

Now we know that 𝐡𝐢 is equal to 𝐡𝐺 since they’re the sides of a square and similarly that 𝐴𝐡 is equal to 𝐡𝐹. And since our triangles have two sides of the same length, including the same angle measure, this is enough to show that the triangles are congruent.

Next, we know that the area of a triangle is half the length of the triangle’s base multiplied by the triangle’s perpendicular height. Applying this to our triangle 𝐺𝐡𝐴, which is an obtuse triangle, and choosing 𝐡𝐺 as the base, we note that 𝐷𝐡 is actually our perpendicular height. So the area of triangle 𝐴𝐡𝐺 is one over two multiplied by 𝐡𝐺 times 𝐡𝐷. That’s one over two times the base times the height. Now we know that 𝐡𝐺 times 𝐡𝐷 is the area of the rectangle 𝐡𝐷𝐾𝐺. Hence, the area of our triangle 𝐴𝐡𝐺 is one over two times the area of the rectangle 𝐡𝐷𝐾𝐺.

Similarly, we can see choosing the base of triangle 𝐡𝐢𝐹 to be 𝐡𝐹 that the area of our triangle 𝐡𝐢𝐹 is one over two multiplied by 𝐡𝐹 times 𝐴𝐡. That’s where 𝐴𝐡 is the perpendicular height of the triangle. And we know that in this case 𝐡𝐹 multiplied by 𝐴𝐡 is the area of the square 𝐴𝐸𝐹𝐡 so that the area of our triangle 𝐡𝐢𝐹 is one over two times the area of the square 𝐴𝐸𝐹𝐡. Now, recalling that our two triangles 𝐴𝐡𝐺 and 𝐡𝐢𝐹 are congruent, they must have the same area. And this means that the square 𝐴𝐸𝐹𝐡 must have the same area as the rectangle 𝐡𝐷𝐾𝐺.

So now, if we equate the expressions for their areas, that’s 𝐡𝐴 squared is equal to 𝐡𝐷 multiplied by 𝐷𝐾. 𝐡𝐴 squared is the area of our square, and 𝐡𝐷 multiplied by 𝐷𝐾 is the area of the rectangle, noting further that the side 𝐷𝐾 has the same length as the side 𝐢𝐻 and that 𝐢𝐻 has the same length as 𝐢𝐡. Since they’re the sides of a square, we have that 𝐡𝐴 squared is equal to 𝐡𝐷 multiplied by 𝐡𝐢. Making a note of this, in the same way, we can show that 𝐢𝐴 squared is equal to 𝐢𝐷 multiplied by 𝐢𝐡. So now making some space, we can summarize our results in the Euclidean theorem. This says if triangle 𝐴𝐡𝐢 is a right triangle at 𝐴 with a projection to 𝐷 as shown, then 𝐡𝐴 squared is equal to 𝐡𝐷 multiplied by 𝐡𝐢 and 𝐢𝐴 squared is equal to 𝐢𝐷 multiplied by 𝐢𝐡.

We can also use the Euclidean theorem, together with the Pythagorean theorem, to show another useful result. Now recalling that the Pythagorean theorem tells us that the hypotenuse of a right angle triangle squared is equal to the sum of the squares of the other two sides. For our triangle 𝐴𝐡𝐷, this gives us 𝐴𝐷 squared plus 𝐡𝐷 squared is 𝐴𝐡 squared. And subtracting 𝐡𝐷 squared from both sides, we have 𝐴𝐷 squared is 𝐴𝐡 squared minus 𝐡𝐷 squared. Now, using our Euclidean theorem, we substitute 𝐴𝐡 squared, that is, 𝐡𝐴 squared, which is 𝐡𝐷 multiplied by 𝐡𝐢. So we have 𝐴𝐷 squared is equal to 𝐡𝐷 multiplied by 𝐡𝐢 minus 𝐡𝐷 squared. And taking out a common factor of 𝐡𝐷, on our right-hand side, we have 𝐡𝐷 multiplied by 𝐡𝐢 minus 𝐡𝐷.

And noting that the side length 𝐡𝐢 is the same as 𝐡𝐷 plus 𝐢𝐷, subtracting 𝐡𝐷 from both sides gives us 𝐢𝐷 is equal to 𝐡𝐢 minus 𝐡𝐷. And since 𝐡𝐢 minus 𝐡𝐷 is the second term on our right-hand side, we have 𝐴𝐷 squared is 𝐡𝐷 multiplied by 𝐢𝐷. And so our corollary to the Euclidean theorem is that if 𝐴𝐡𝐢 is a right triangle at 𝐴 with projection to 𝐷 as shown, then 𝐴𝐷 squared is 𝐡𝐷 multiplied by 𝐢𝐷.

It’s worth noting that there’s an alternative and equivalent way of looking at these two results. This is the Euclidean leg rule, where again we have a right triangle 𝐴𝐡𝐢 with right angle at 𝐴. And this is projected onto the hypotenuse 𝐢𝐡 to the point 𝐷. And this tells us if the leg and the part are as shown, then the hypotenuse divided by the leg is equal to the leg divided by the part. That is, 𝐡𝐢 over 𝐡𝐴 is equal to 𝐡𝐴 over 𝐡𝐷. And if we multiply both sides by 𝐡𝐴 and 𝐡𝐷, this gives us the first part of our Euclidean theorem: 𝐡𝐢 multiplied by 𝐡𝐷 is 𝐡𝐴 squared.

Now, if we change our leg and part to the triangle 𝐴𝐷𝐢 as opposed to 𝐴𝐷𝐡, again we have hypotenuse over leg is equal to leg over part. That is, 𝐡𝐢 over 𝐢𝐴 is equal to 𝐢𝐴 over 𝐢𝐷. And we have the second part of our Euclidean theorem: 𝐡𝐢 multiplied by 𝐢𝐷 is 𝐢𝐴 squared. And similarly, for our second result, the corollary, also called the altitude rule, if the altitude is the perpendicular height 𝐴𝐷 and the hypotenuse is split into left and right parts, that’s 𝐢𝐷 and 𝐷𝐡, we have left over altitude is equal to altitude over right. And this gives us our corollary 𝐢𝐷 multiplied by 𝐡𝐷 is 𝐴𝐷 squared. And what this tells us is that the altitude or height of the right triangle squared is the product of the lengths of the segments of the hypotenuse when it’s split by the altitude.

So now let’s look at an example of applying the Euclidean theorem to find a missing side length in a right triangle.

Find the length of the line segment 𝐴𝐷.

To find the length 𝐴𝐷, we first note that 𝐷 is the projection of 𝐡 onto 𝐴𝐢 and that the triangle 𝐴𝐡𝐢 is a right triangle at 𝐡. And now recalling the Euclidean theorem, this tells us that 𝐴𝐡 squared is 𝐴𝐷 multiplied by 𝐴𝐢. That is, the side length 𝐴𝐡 squared is the product of the segment 𝐴𝐷 with the length of the hypotenuse 𝐴𝐢. We’re given that 𝐴𝐡 is 34 centimeters and that the hypotenuse 𝐴𝐢 is 40 centimeters. And substituting these values into the Euclidean theorem, this gives us 34 squared is 𝐴𝐷 multiplied by 40. Dividing through by 40 and rearranging, this gives us 𝐴𝐷 is 34 squared over 40, that is, 1156 divided by 40, which is 28.9. The line segment 𝐴𝐷 is therefore 28.9 centimeters.

Let’s now see another example of applying the Euclidean theorem to determine the length of a side in a right triangle.

Determine the length of line segment 𝐴𝐷.

We note that 𝐷 is the projection of 𝐴 onto the line 𝐡𝐢 and that the triangle 𝐴𝐡𝐢 is a right triangle at 𝐴. And we recall that the corollary to the Euclidean theorem, that is, the altitude rule, tells us that 𝐴𝐷 squared is equal to 𝐡𝐷 multiplied by 𝐢𝐷. This tells us that the altitude or height of the right triangle squared is the product of the lengths of the segments of the hypotenuse when it is split by the altitude.

We’re given that the two segments are 2.5 centimeters, that’s 𝐢𝐷, and 6.4 centimeters, that’s 𝐡𝐷. And substituting these values in, we have 𝐴𝐷 squared is 6.4 multiplied by 2.5. This evaluates to 16, so 𝐴𝐷 squared is 16. Now, taking the square root on both sides of the equation, noting that 𝐴𝐷 is a length and so is nonnegative, we get 𝐴𝐷 is the square root of 16, which is four. The length of 𝐴𝐷 is therefore four centimeters.

In our next example, we’ll apply both parts of the Euclidean theorem to determine a missing side length in a right triangle.

In the figure shown, if 𝑋𝐿 is equal to 40 and π‘ŒπΏ is 30, what is the length of π‘Œπ‘?

We note first that 𝐿 is the projection of π‘Œ onto 𝑋𝑍. And the triangle π‘‹π‘Œπ‘ is a right triangle at π‘Œ. We can then recall that the corollary to the Euclidean theorem tells us that π‘ŒπΏ squared is 𝐿𝑍 multiplied by 𝑋𝐿. Now, since we know the lengths of 𝑋𝐿 and π‘ŒπΏ, that is, 40 and 30 centimeters, respectively, this will allow us to find 𝐿𝑍. We can then apply the Euclidean theorem to find π‘Œπ‘. So now, substituting in the given lengths to the corollary of the Euclidean theorem, which is also known as the altitude rule, we have 30 squared is 𝐿𝑍 multiplied by 40. Dividing through by 40 and rearranging, this gives us 𝐿𝑍 is equal to 30 squared divided by 40, that is, 900 divided by 40, which is 22.5.

So making a note of this and clearing some space, we can now use the Euclidean theorem to note that π‘Œπ‘ squared is equal to 𝐿𝑍 multiplied by 𝑍𝑋. Remember, 𝐿𝑍 is the shortest segment of the hypotenuse, which is 22.5 centimeters, and 𝑍𝑋 is the length of the hypotenuse of the triangle π‘‹π‘Œπ‘. We know that the length of the hypotenuse 𝑍𝑋 is the sum of the two segments, that is, 40 plus 22.5. And that’s equal to 62.5. Now, substituting these values into our Euclidean theorem, that’s π‘Œπ‘ squared is equal to 𝐿𝑍 multiplied by 𝑍𝑋, we have 𝐿𝑍 is 22.5 multiplied by 𝑍𝑋, which is 62.5. And this evaluates to 1406.25. And now taking the positive square root on both sides, since π‘Œπ‘ is a length and so must always be positive, we have that π‘Œπ‘ is 37.5 centimeters.

It’s worth noting that we could also have found this length using the Pythagorean theorem with the right triangle π‘ŒπΏπ‘. This gives us π‘Œπ‘ squared is equal to 𝐿𝑍 squared plus π‘ŒπΏ squared. That is, π‘Œπ‘ squared is 22.5 squared plus 30 squared, which is 1406.25 as before. And so again, taking the positive square root, we have π‘Œπ‘ is 37.5. And so the length π‘Œπ‘ is 37.5 centimeters.

In our final example, we’ll apply the Euclidean theorem and the Pythagorean theorem to determine the area of the right triangle from a given diagram.

Calculate the area of triangle 𝐴𝐡𝐷.

We begin by recalling that the area of a triangle is half the length of the triangle’s base multiplied by the triangle’s perpendicular height. And since 𝐴𝐡𝐷 is a right triangle at 𝐷, we can choose 𝐡𝐷 as the base and then 𝐴𝐷 as the perpendicular height. The area of the triangle is then one over two times 𝐡𝐷 multiplied by 𝐴𝐷. We can find the length 𝐴𝐷 using the Euclidean theorem and then the length 𝐡𝐷 using the Pythagorean theorem.

If we first note that 𝐷 is the projection of 𝐡 onto the side 𝐴𝐢 and that the triangle 𝐴𝐡𝐢 is a right triangle at 𝐡, then the Euclidean theorem tells us that 𝐴𝐡 squared is equal to 𝐴𝐢 multiplied by 𝐴𝐷. Substituting now 𝐴𝐡 is 44 centimeters and 𝐴𝐢 is 55 centimeters, we have 44 squared is 55 multiplied by 𝐴𝐷. And dividing through by 55, we have 44 squared over 55 is 𝐴𝐷, which evaluates to 35.2 centimeters. And we now have the lengths of two sides of our right triangle 𝐴𝐡𝐷. And so we can use the Pythagorean theorem to find a third length, that is, the length 𝐡𝐷.

Applied to our triangle, the Pythagorean theorem tells us that 𝐴𝐡 squared is 𝐴𝐷 squared plus 𝐡𝐷 squared. And substituting 𝐴𝐡 is 44 centimeters and 𝐴𝐷 is 35.2 centimeters, we have 44 squared is 35.2 squared plus 𝐡𝐷 squared. And subtracting 35.2 squared from both sides and rearranging, this gives us 𝐡𝐷 squared is 44 squared minus 35.2 squared. Our right-hand side evaluates to 696.96. And taking the positive square root on both sides, since 𝐡𝐷 is a length, that’s the square root of 696.96, which is equal to 26.4. So 𝐡𝐷 is 26.4 centimeters.

So now, we can substitute 𝐴𝐷 is 35.2 centimeters and 𝐡𝐷 is 26.4 centimeters into our formula for the area. This gives us the area of triangle 𝐴𝐡𝐷 is one over two times 26.4 multiplied by 35.2. This evaluates to 464.64, and hence the area of triangle 𝐴𝐡𝐷 is 464.64 square centimeters.

Let’s now complete this video by recapping some of the key points we’ve covered. The Euclidean theorem tells us that if 𝐴𝐡𝐢 is a right triangle at 𝐴 with projection to 𝐷 on the line 𝐢𝐡, then 𝐡𝐴 squared is 𝐡𝐷 multiplied by 𝐡𝐢 and 𝐢𝐴 squared is 𝐢𝐷 multiplied by 𝐢𝐡. There is a useful corollary to the Euclidean theorem that gives us 𝐴𝐷 squared is equal to 𝐡𝐷 multiplied by 𝐢𝐷. And this is also called the altitude rule. Sometimes we need to apply both the Euclidean theorem and its corollary. We may also need to apply these in conjunction with the Pythagorean theorem to solve problems.

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