# 1.6: The Pythagorean Theorem for Area and Perimeter

Find missing sides to calculate the area

### Applications of the Pythagorean Theorem

#### Find the Height of an Isosceles Triangle

One way to use The Pythagorean Theorem is to find the height of an isosceles triangle (see Example 1).

#### Prove the Distance Formula

Another application of the Pythagorean Theorem is the Distance Formula. We will prove it here.

Let’s start with point $$A(x_1, y_1)$$ and point $$B(x_2, y_2)$$. We will call the distance between A and B, d.

Draw the vertical and horizontal lengths to make a right triangle.

Now that we have a right triangle, we can use the Pythagorean Theorem to find the hypotenuse, d.

\begin{aligned} d^2&=\sqrt{(x_1−x_2)^2+(y_1−y_2)^2} \\ d&=(x_1−x_2)^2+(y_1−y_2)^2 \end{aligned}

Distance Formula: The distance between $$A(x_1,y_1)$$ and $$B(x_2,y_2)$$ is

$$d=\sqrt{(x_1−x_2)^2+(y_1−y_2)^2}$$.

#### Classify a Triangle as Acute, Right, or Obtuse

We can extend the converse of the Pythagorean Theorem to determine if a triangle is an obtuse or acute triangle.

Acute Triangles: If the sum of the squares of the two shorter sides in a right triangle is greater than the square of the longest side, then the triangle is acute.

For $$b<c$$ and $$a<c$$, if $$a^2+b^2>c^2$$, then the triangle is acute.

Obtuse Triangles: If the sum of the squares of the two shorter sides in a right triangle is less than the square of the longest side, then the triangle is obtuse.

For $$b<c$$ and $$a<c$$, if $$a^2+b^2<c^2$$, then the triangle is obtuse.

What if you were given an equilateral triangle in which all the sides measured 4 inches? How could you use the Pythagorean Theorem to find the triangle's altitude?

Example $$\PageIndex{1}$$

What is the height of the isosceles triangle?

Solution

Draw the altitude from the vertex between the congruent sides, which will bisect the base.

\begin{aligned} 7^2+h^2 &=9^2 \\ 49+h^2&=81 \\ h^2&=32 \\ h&=\sqrt{32}=\sqrt{16 \cdot 2}=4\sqrt{2}\end{aligned}

Example $$\PageIndex{2}$$

Find the distance between $$(1, 5)$$ and $$(5, 2)$$.

Solution

Make $$A(1,5)$$ and $$B(5,2)$$. Plug into the distance formula.

\begin{aligned} d&=\sqrt{(1−5)^2+(5−2)^2} \\ &=\sqrt{(−4)^2+(3)^2}\\ &=\sqrt{16+9}=\sqrt{25}=5 \end{aligned}

Just like the lengths of the sides of a triangle, distances are always positive.

Example $$\PageIndex{3}$$

Graph $$A(−4,1)$$, $$B(3,8)$$, and $$C(9,6)$$. Determine if $$\Delta ABC$$ is acute, obtuse, or right.

Solution

Use the distance formula to find the length of each side.

\begin{aligned} AB&=\sqrt{(−4−3)^2+(1−8)^2}=\sqrt{49+49}=\sqrt{98} \\ BC&=\sqrt{(3−9)^2+(8−6)^2}=\sqrt{36+4}=\sqrt{40} \\ AC&=\sqrt{(−4−9)^2+(1−6)^2}=\sqrt{169+25}=\sqrt{194}\end{aligned}

Plug these lengths into the Pythagorean Theorem.

\begin{aligned} (\sqrt{98})^2+(\sqrt{40})^2 &? (\sqrt{194})^2 \\ 98+40 &? 194\\ 138 &<194 \end{aligned}

$$\Delta ABC$$ is an obtuse triangle.

For Examples 4 and 5, determine if the triangles are acute, right or obtuse.

Example $$\PageIndex{4}$$

Set the longest side to $$c$$.

Solution

\begin{aligned} 15^2+14^2 &? 21^2 \\ 225+196 &? 441 \\ 421 &<441\end{aligned}

The triangle is obtuse.

Example $$\PageIndex{5}$$

Set the longest side to $$c$$.

A triangle with side lengths 5, 12, 13.

Solution

$$5^2+12^2=13^2$$ so this triangle is right.

## Review

Find the height of each isosceles triangle below. Simplify all radicals.

Find the length between each pair of points.

1. (-1, 6) and (7, 2)
2. (10, -3) and (-12, -6)
3. (1, 3) and (-8, 16)
4. What are the length and width of a 42” HDTV? Round your answer to the nearest tenth.
5. Standard definition TVs have a length and width ratio of 4:3. What are the length and width of a 42” Standard definition TV? Round your answer to the nearest tenth.

Determine whether the following triangles are acute, right or obtuse.

1. 7, 8, 9
2. 14, 48, 50
3. 5, 12, 15
4. 13, 84, 85
5. 20, 20, 24
6. 35, 40, 51
7. 39, 80, 89
8. 20, 21, 38
9. 48, 55, 76

Graph each set of points and determine whether $$\Delta ABC$$ is acute, right, or obtuse, using the distance formula.

1. $$A(3,−5), B(−5,−8), C(−2,7)$$
2. $$A(5,3),B(2,−7),C(−1,5)$$
3. $$A(1,6),B(5,2),C(−2,3)$$
4. $$A(−6,1),B(−4,−5),C(5,−2)$$