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Lesson Explainer: Right Triangle Altitude Theorem Mathematics • 11th Grade

In this explainer, we will 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 will allow us to determine the lengths of sides in a right triangle given two of the lengths.

Let’s start by deriving the Euclidean theorem. To do this, we first construct a right triangle 𝐴𝐡𝐢 with the right angle at 𝐴. We then project 𝐴 onto 𝐢𝐡 and call this point 𝐷 as shown.

We now add to the diagram three squares given by each side of 𝐴𝐡𝐢 and label the vertices of these squares as shown.

We now want to project 𝐴 onto 𝐺𝐻 and add the lines 𝐢𝐹 and 𝐴𝐺 to the diagram as shown.

We now want to show that △𝐢𝐡𝐹 and △𝐺𝐡𝐴 are congruent. We do this by first noting that π‘šβˆ πΆπ΅πΉ=π‘šβˆ π΄π΅πΊ since these are both right angles added to ∠𝐴𝐡𝐢. We know that 𝐡𝐢=𝐡𝐺 and that 𝐴𝐡=𝐡𝐹. This gives us the following.

This is enough to show 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 △𝐺𝐡𝐴, we choose 𝐡𝐺 as the base and we note that 𝐷𝐡 is the perpendicular height as shown.

Thus, area△𝐴𝐡𝐺=12𝐡𝐺×𝐡𝐷.

We know that (𝐡𝐺)Γ—(𝐡𝐷) is the area of 𝐡𝐷𝐾𝐺. So, areaarea△𝐴𝐡𝐺=12(𝐡𝐷𝐾𝐺).

Similarly, we can see that area△𝐡𝐢𝐹=12𝐡𝐹×𝐴𝐡.

We also note that 𝐡𝐹×𝐴𝐡=(𝐴𝐸𝐹𝐡).area

So, areaarea△𝐡𝐢𝐹=12(𝐴𝐸𝐹𝐡).

Finally, since these triangles are congruent, they have the same area, which means that square 𝐴𝐸𝐹𝐡 has the same area as rectangle 𝐡𝐷𝐾𝐺. We can then equate the expressions for their areas: 𝐴𝐡=𝐡𝐷×𝐷𝐾.

We note that 𝐷𝐾=𝐢𝐻 and that 𝐢𝐻=𝐡𝐢, so we can rewrite this as 𝐴𝐡=𝐡𝐷×𝐡𝐢.

We can show a similar result for 𝐴𝐢=𝐢𝐷×𝐡𝐢.

We can summarize the result we have shown as follows.

Theorem: Euclidean Theorem

In any right triangle, the area of the square on a side adjacent to the right angle is equal to the area of the rectangle whose dimensions are the length of the projection of this side on the hypotenuse and the length of the hypotenuse.

In general, if 𝐴𝐡𝐢 is a right triangle at 𝐴 with a projection to 𝐷 as shown, then 𝐴𝐡=𝐡𝐷×𝐡𝐢,𝐴𝐢=𝐢𝐷×𝐡𝐢.

We can use the Euclidean theorem together with the Pythagorean theorem to show another useful result. First, the Pythagorean theorem gives us 𝐴𝐷=π΄π΅βˆ’π΅π·.

Next, we can use the Euclidean theorem to substitute 𝐴𝐡=π΅π·Γ—π΅πΆοŠ¨ into this equation. We get 𝐴𝐷=π΅π·Γ—π΅πΆβˆ’π΅π·.

We can then take out the shared factor of 𝐡𝐷 on the right-hand side of the equation to get 𝐴𝐷=𝐡𝐷(π΅πΆβˆ’π΅π·).

Finally, we know that 𝐡𝐢=𝐡𝐷+𝐢𝐷, so π΅πΆβˆ’π΅π·=𝐢𝐷. Hence, 𝐴𝐷=𝐡𝐷×𝐢𝐷.

We can write this result formally as follows.

Theorem: Corollary to the Euclidean Theorem

If 𝐴𝐡𝐢 is a right triangle at 𝐴 with projection to 𝐷 as shown, then 𝐴𝐷=𝐡𝐷×𝐢𝐷.

Let’s now see some examples of applying the Euclidean theorem to find missing side lengths in right triangles.

Example 1: Finding a Side Length in a Right Triangle Using the Euclidean Theorem

Find the length of 𝐴𝐷.

Answer

We note that 𝐷 is the projection of 𝐡 onto 𝐴𝐢 and that △𝐴𝐡𝐢 is a right triangle at 𝐡. We can then recall that the Euclidean theorem tells us that 𝐴𝐡=𝐴𝐷×𝐴𝐢.

Substituting in the given lengths, we have 34=𝐴𝐷×40.

We can then divide the equation through by 40 to get 𝐴𝐷=3440=28.9.cm

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

Example 2: Finding the Altitude Using the Hypotenuse Divided into Two Segments

Determine the length of 𝐴𝐷.

Answer

We note that 𝐷 is the projection of 𝐴 onto 𝐡𝐢 and that △𝐴𝐡𝐢 is a right triangle at 𝐴. We can then recall that the corollary of the Euclidean theorem tells us that 𝐴𝐷=𝐢𝐷×𝐡𝐷.

Substituting in the given lengths gives 𝐴𝐷=2.5Γ—6.4=16.

Taking the square root of both sides of the equation, noting that 𝐴𝐢 is a length and so is nonnegative, we get 𝐴𝐷=√16=4.cm

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

Example 3: Finding an Unknown Length Using Both Parts of the Euclidean Theorem

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

Answer

We note that 𝐿 is the projection of π‘Œ onto 𝑋𝑍 and that β–³π‘‹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=𝐿𝑍×𝑋𝐿.

Since we know the lengths of 𝑋𝐿 and π‘ŒπΏ, this will allow us to find 𝐿𝑍. Then, we can apply the Euclidean theorem to find π‘Œπ‘.

Substituting in the given lengths into the corollary of the Euclidean theorem gives us 30=𝐿𝑍×40.

We divide the equation through by 40 to see that 𝐿𝑍=3040=22.5.

We can now use the Euclidean theorem to note that π‘Œπ‘=𝐿𝑍×𝑍𝑋.

We know that 𝑍𝑋=𝑋𝐿+𝐿𝑍=22.5+40=62.5.

Substituting 𝐿𝑍=22.5 and 𝑍𝑋=62.5 gives π‘Œπ‘=22.5Γ—62.5=1406.25.

Taking the square root of both sides of the equation, noting that π‘Œπ‘ is a length and so is nonnegative, we get π‘Œπ‘=√1406.25=37.5.

It is worth noting that we could also have found this length using the Pythagorean theorem. Applying the Pythagorean theorem to right triangle π‘ŒπΏπ‘, we get π‘Œπ‘=𝐿𝑍+π‘ŒπΏ.

Substituting 𝐿𝑍=22.5 and π‘ŒπΏ=30 gives π‘Œπ‘=22.5+30=1406.25.

Hence, π‘Œπ‘=√1406.25=37.5.

In our next example, we will apply the Euclidean theorem and Pythagorean theorem to determine a missing length in a right triangle.

Example 4: Finding an Unknown Length Using the Euclidean and Pythagorean Theorems

From the figure, determine the length of 𝐡𝐷. If necessary, round your answer to the nearest hundredth.

Answer

We can start by applying the Pythagorean theorem to right triangle 𝐴𝐡𝐢. This gives 𝐴𝐢=𝐴𝐡+𝐡𝐢.

Since we know the lengths of 𝐡𝐢 and 𝐴𝐡, this will allow us to find 𝐴𝐢. We can then apply the Euclidean theorem to find 𝐢𝐷. Finally, we can then apply the Pythagorean theorem to △𝐡𝐷𝐢 to find 𝐡𝐷.

Substituting 𝐴𝐡=8 and 𝐡𝐢=15 gives us 𝐴𝐢=8+15=289.

Taking the square root of both sides of the equation gives us 𝐴𝐢=√289=17.cm

We note that 𝐷 is the projection of 𝐡 onto 𝐴𝐢 and that △𝐴𝐡𝐢 is a right triangle at 𝐡. We can then recall that the Euclidean theorem tells us that 𝐢𝐡=𝐢𝐷×𝐴𝐢.

Substituting 𝐢𝐡=15 and 𝐴𝐢=17 gives 15=𝐢𝐷×17.

Dividing the equation through by 17 then yields 𝐢𝐷=1517=22517.

Finally, we apply the Pythagorean theorem to triangle 𝐡𝐷𝐢. This gives 𝐡𝐷=π΅πΆβˆ’πΆπ·.

Substituting 𝐢𝐷=22517 and 𝐡𝐢=15 gives us 𝐡𝐷=15βˆ’ο€Ό22517=49.82….

We then take the square root of both sides of the equation to get 𝐡𝐷=√49.826…=7.058….cm

Rounding to the nearest hundredth gives us 7.06 cm.

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

Example 5: Finding the Area of a Triangle Using the Right Triangle Theorems

Calculate the area of △𝐴𝐡𝐷.

Answer

We start by recalling that the area of a triangle is half the length of the triangle’s base multiplied by the triangle’s perpendicular height. Since △𝐴𝐡𝐷 is a right triangle at 𝐷, we can choose 𝐡𝐷 as the base and then 𝐴𝐷 is the perpendicular height, giving us area△𝐴𝐡𝐷=12𝐡𝐷×𝐴𝐷.

We can find the length of 𝐴𝐷 using the Euclidean theorem and then the length of 𝐡𝐷 using the Pythagorean theorem. First, we note that 𝐷 is the projection of 𝐡 onto 𝐴𝐢 and that △𝐴𝐡𝐢 is a right triangle at 𝐡. The Euclidean theorem then tells us that 𝐴𝐡=𝐴𝐢×𝐴𝐷.

We substitute 𝐴𝐡=44cm and 𝐴𝐢=55cm to get 44=55×𝐴𝐷.

We then divide the equation through by 55, giving us 𝐴𝐷=4455=35.2.cm

We now have the lengths of two sides of right triangle 𝐴𝐡𝐷. We can find the third length by applying the Pythagorean theorem, which gives 𝐴𝐡=𝐴𝐷+𝐡𝐷.

Substituting 𝐴𝐡=44cm and 𝐴𝐷=35.2cm gives 44=35.2+𝐡𝐷.

We can then rearrange and simplify the equation: 𝐡𝐷=44βˆ’35.2=696.96.

We then take the square root of both sides of the equation, noting that 𝐡𝐷 is a length and so is nonnegative, which gives 𝐡𝐷=√696.96=26.4.cm

We now substitute 𝐴𝐷=35.2cm and 𝐡𝐷=26.4cm into the formula for the area of △𝐴𝐡𝐷 to get areacm△𝐴𝐡𝐷=12(26.4)Γ—(35.2)=464.64.

In our final example, we will apply both the Euclidean and Pythagorean theorems to determine the length of a missing side in a given diagram.

Example 6: Finding an Unknown Length in a Square Divided into Triangles Using Right Triangle Theorems

Find the length of 𝐴𝐹, approximating the result to the nearest hundredth.

Answer

We start by applying the Pythagorean theorem to right triangle 𝐡𝐴𝐷. This gives 𝐡𝐷=𝐴𝐡+𝐴𝐷.

Since we know the values of 𝐴𝐷 and 𝐴𝐡, we can use this to find 𝐡𝐷. We can then use the Euclidean theorem and the values of 𝐡𝐷 and 𝐴𝐷 to find 𝐹𝐷. Finally, we can apply the Pythagorean theorem to △𝐴𝐹𝐷 to find 𝐴𝐹.

Substituting 𝐴𝐡=33 and 𝐴𝐷=56 into the above equation gives us 𝐡𝐷=33+56=4225.

Taking the square root of both sides of the equation gives us 𝐡𝐷=√4225=65.cm

We note that 𝐹 is the projection of 𝐴 onto 𝐡𝐷 and that △𝐡𝐴𝐷 is a right triangle at 𝐴. We can then recall that the Euclidean theorem tells us that 𝐴𝐷=𝐡𝐷×𝐹𝐷.

Substituting 𝐴𝐷=56 and 𝐡𝐷=65 gives 56=65×𝐹𝐷.

Dividing the equation through by 65 then yields 𝐹𝐷=5665=313665.

Finally, we apply the Pythagorean theorem to triangle 𝐴𝐹𝐷. This gives 𝐴𝐹=π΄π·βˆ’πΉπ·.

Substituting 𝐹𝐷=313665 and 𝐴𝐷=56 gives us 𝐴𝐹=56βˆ’ο€Ό313665=808.308….

We then take the square root of both sides of the equation to get 𝐴𝐹=√808.308…=28.430….cm

Rounding to the nearest hundredth gives us 28.43 cm.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The Euclidean theorem tells us that if 𝐴𝐡𝐢 is a right triangle at 𝐴 with projection to 𝐷 as shown, then 𝐴𝐡=𝐡𝐷×𝐡𝐢,𝐴𝐢=𝐢𝐷×𝐡𝐢.
  • There is a useful corollary to the Euclidean theorem that we find by applying the Pythagorean; we can also show that 𝐴𝐷=𝐡𝐷×𝐢𝐷.
  • Sometimes we will need to apply both the Euclidean theorem and its corollary. We may also need to apply these results in conjunction with the Pythagorean theorem to solve problems.

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