### Area and Perimeter of Triangles

If we take parallelogram and cut it in half, along a diagonal, we would have two congruent triangles. Therefore, the formula for the area of a triangle is the same as the formula for area of a parallelogram, but cut in half.

The **area of a triangle** is \begin{align*}A=\frac{1}{2} bh\end{align*} or \begin{align*}A=\frac{bh}{2}\end{align*}. In the case that the triangle is a right triangle, then the height and base would be the legs of the right triangle. If the triangle is an obtuse triangle, the altitude, or height, could be outside of the triangle.

#### Calculating the Area

Find the area of the triangle.

This is an obtuse triangle. To find the area, we need to find the height of the triangle. We are given the two sides of the small right triangle, where the hypotenuse is also the short side of the obtuse triangle. From these values, we see that the height is 4 because this is a 3-4-5 right triangle. The area is \begin{align*}A=\frac{1}{2} (4)(7)=14 \ units^2\end{align*}.

#### Calculating the Perimeter

Find the perimeter of the triangle from the previous Example.

To find the perimeter, we would need to find the longest side of the obtuse triangle. If we used the dotted lines in the picture, we would see that the longest side is also the hypotenuse of the right triangle with legs 4 and 10. Use the Pythagorean Theorem.

\begin{align*}4^2+10^2 &= c^2\\ 16+100 &= c^2\\ c &= \sqrt{116} \approx 10.77 \qquad \text{The perimeter is} \ 7 + 5 + 10. 77 = 22.77 \ units\end{align*}

#### Finding the Area given Length and Height

Find the area of a triangle with base of length 28 cm and height of 15 cm.

The area is \begin{align*}\frac{1}{2}(28)(15)=210 \ cm^2\end{align*}.

### Examples

Use the triangle to answer the following questions.

#### Example 1

Find the height of the triangle.

Use the Pythagorean Theorem to find the height.

\begin{align*} 8^2+h^2&=17^2\\ h^2&=225\\ h&=15 \ in\end{align*}

#### Example 2

Find the perimeter.

We need to find the hypotenuse. Use the Pythagorean Theorem again.

\begin{align*} (8+24)^2+15^2&=h^2\\ h^2&=1249\\ h&\approx 35.3 \ in\end{align*}

#### Example 3

Find the area.

The perimeter is \begin{align*}24+35.3+17 \approx 76.3 \ in\end{align*}.

3. The area is \begin{align*}\frac{1}{2}(24)(15)=180 \ in^2\end{align*}.

### Review

Use the triangle to answer the following questions.

- Find the height of the triangle by using the geometric mean.
- Find the perimeter.
- Find the area.

Find the area of the following shape.

- What is the height of a triangle with area \begin{align*}144 \ m^2\end{align*} and a base of 24 m?

For problems 6 and 7 find the height and area of the equilateral triangle with the given perimeter.

- Perimeter 18 units.
- Perimeter 30 units.
- Generalize your results from problems 6 and 7 into a formula to find the height and area of an equilateral triangle with side length \begin{align*}x\end{align*}.

Find the area of each triangle.

- Find the area of a triangle with a base of 10 in and a height of 12 in.
- Find the area of a triangle with a base of 5 in and a height of 3 in.
- An equilateral triangle with a height of \begin{align*}6\sqrt{3}\end{align*} units.
- A 45-45-90 triangle with a hypotenuse of \begin{align*}5\sqrt{2}\end{align*} units.
- A 45-45-90 triangle with a leg of 12 units.
- A 30-60-90 triangle with a hypotenuse of 24 units.
- A 30-60-90 triangle with a short leg of 5 units.

### Review (Answers)

To view the Review answers, open this PDF file and look for section 10.3.