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# Lengths of Triangle Sides Using the Pythagorean Theorem

## Discover, Geometrically prove, and apply the theorem

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Pythagorean Theorem and its Converse

Mr. Aubel wants to rope off half of his rectangular garden plot to keep the deer out. He will run the rope around the outside of the garden and diagonally down the center to form a right triangle. The garden measures 5 yards by 8 yards. How many full yards of rope does Mr. Aubel need?

### Pythagorean Theorem and Its Inverse

The Pythagorean Theorem refers to the relationship between the lengths of the three sides in a right triangle. It states that if a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*} are the legs of the right triangle and c\begin{align*}c\end{align*} is the hypotenuse, then a2+b2=c2\begin{align*}a^2+b^2=c^2 \end{align*}. For example, the lengths 3, 4, and 5 are the sides of a right triangle because 32+42=52(9+16=25)\begin{align*}3^2+4^2=5^2 (9+16 = 25)\end{align*}. Keep in mind that c\begin{align*}c\end{align*} is always the longest side.

The converse of this statement is also true. If, in a triangle, c\begin{align*}c\end{align*} is the length of the longest side and the shorter sides have lengths a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}, and a2+b2=c2\begin{align*}a^2+b^2=c^2\end{align*}, then the triangle is a right triangle.

#### Proof of Pythagorean Theorem

There are many proofs of the Pythagorean Theorem and here is one of them. We will be using the concept that the area of a figure is equal to the sum of the areas of the smaller figures contained within it and algebra to derive the Pythagorean Theorem.

Using the figure below (a square with a smaller square inside), first write two equations for its area, one using the lengths of the sides of the outer square and one using the sum of the areas of the smaller square and the four triangles.

Area 1: (a+b)2=a2+2ab+b2\begin{align*}(a+b)^2=a^2+2ab+b^2\end{align*}

Area 2: c2+4(12ab)=c2+2ab\begin{align*}c^2+4 \left(\frac{1}{2} ab \right)=c^2+2ab\end{align*}

Now, equate the two areas and simplify:

a2+2ab+b2a2+b2=c2+2ab=c2\begin{align*}a^2+2ab+b^2&=c^2+2ab\\ a^2+b^2&=c^2\end{align*}

In a right triangle a=7\begin{align*}a=7\end{align*} and c=25\begin{align*}c=25\end{align*}, let's find the length of the third side.

We can start by substituting what we know into the Pythagorean Theorem and then solve for the unknown side, b\begin{align*}b\end{align*}:

72+b249+b2b2b=252=625=576=24\begin{align*}7^2+b^2 &=25^2 \\ 49+b^2 &=625 \\ b^2 &=576 \\ b &=24\end{align*}

Now, let's find the length of the third side of the triangle below and leave our answer in reduced radical form.

Since we are given the lengths of the two legs, we can plug them into the Pythagorean Theorem and find the length of the hypotenuse.

82+12264+144c2c=c2=c2=208=208=1613=413\begin{align*}8^2+12^2 &=c^2 \\ 64+144 &=c^2 \\ c^2 &=208 \\ c &=\sqrt{208}=\sqrt{16 \cdot 13}=4\sqrt{13}\end{align*}

Finally, let's determine whether a triangle with lengths 21, 28, 35 is a right triangle.

We need to see if these values will satisfy a2+b2=c2\begin{align*}a^2+b^2=c^2\end{align*}. If they do, then a right triangle is formed. So,

212+282352=441+784=1225=1225\begin{align*}21^2+28^2 &=441+784=1225 \\ 35^2 &=1225\end{align*}

Yes, the Pythagorean Theorem is satisfied by these lengths and a right triangle is formed by the lengths 21, 28 and 35.

### Examples

#### Example 1

Earlier, you were asked to find how many yards of rope Mr. Aubel needs.

We are looking for the perimeter of the triangle. We are given the lengths of the sides so we need to find the hypotenuse.

Let's use the Pythagorean Theorem.

52+82=c225+64=c289=c2c=89\begin{align*}5^2 + 8^2 = c^2\\ 25 + 64 = c^2\\ 89 = c^2\\ c = \sqrt {89}\end{align*}

Now to find the perimeter of the triangle, add the lengths of the three sides.

5+8+89=22.43\begin{align*} 5 + 8 + \sqrt{89} = 22.43\end{align*}

Therefore, Mr. Aubel will need 23 yards of rope.

For the given two sides, determine the length of the third side if the triangle is a right triangle.

#### Example 2

a=10\begin{align*}a=10\end{align*} and b=5\begin{align*}b=5\end{align*}

102+52=100+25=125=55\begin{align*}\sqrt{10^2+5^2}=\sqrt{100+25}=\sqrt{125}=5\sqrt{5}\end{align*}

#### Example 3

a=5\begin{align*}a=5\end{align*} and c=13\begin{align*}c=13\end{align*}

13252=16925=144=12\begin{align*}\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12\end{align*}

Use the Pythagorean Theorem to determine if a right triangle is formed by the given lengths.

#### Example 4

16, 30, 34

162+302342=256+900=1156=1156\begin{align*}16^2+30^2 &=256+900=1156 \\ 34^2 &=1156\end{align*} Yes, this is a right triangle.

#### Example 5

9, 40, 42

92+402422=81+1600=1681=1764\begin{align*}9^2+40^2 &=81+1600=1681 \\ 42^2 &=1764\end{align*} No, this is not a right triangle.

#### Example 6

2, 2, 4

This one is tricky, in a triangle the lengths of any two sides must have a sum greater than the length of the third side. These lengths do not meet that requirement so not only do they not form a right triangle, they do not make a triangle at all.

### Review

Find the unknown side length for each right triangle below.

1. a=6,b=8\begin{align*}a=6, b=8\end{align*}
2. b=6,c=14\begin{align*}b=6, c=14\end{align*}
3. a=12,c=18\begin{align*}a=12, c=18\end{align*}

Determine whether the following triangles are right triangles.

Do the lengths below form a right triangle? Remember to make sure that they form a triangle.

1. 3, 4, 5
2. 6, 6, 11
3. 11, 13, 17

Major General James A. Garfield (and former President of the U.S.) is credited with deriving this proof of the Pythagorean Theorem using a trapezoid. Follow the steps to recreate his proof.

1. Find the area of the trapezoid using the trapezoid area formula: A=12(b1+b2)h\begin{align*}A=\frac{1}{2}(b_1+b_2)h\end{align*}
2. Find the sum of the areas of the three right triangles in the diagram.
3. The areas found in the previous two problems should be the same value. Set the expressions equal to each other and simplify to get the Pythagorean Theorem.

To see the Review answers, open this PDF file and look for section 13.1.

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