TIE - Area of a Triangle Lesson

Area of a Triangle

In previous math courses you've learned how to find the area of triangles using the formula:

equation image indicator $LaTeX: A=\frac{1}{2}bh$ $A=\frac{1}{2}bh$

triangle with height of 10 in and base of 8 in.
A=½bh
A=½(10)(8)
A=40in sq triangle with height of 7 cm and base of 23 in.
A=½bh
A=½(13)(7)
A=45.4 cm sq

But, what if you are given an oblique triangle and you know two sides and their included angle as shown below:

oblique triangle with sides of 12 in, 85°, and base of 8 in.

You have a base length of 8 inches, but you do not know the height. Well imagine that we drew an altitude from equation image indicator $LaTeX: \angle A$ $\angle A$ , down to side $LaTeX: \overline{BC}$ $\overline{BC}$ .

oblique triangle with sides of 12 in, 85°, and base of 8 in. with height denoted

Consider the right triangle formed and the trigonometric ratio between equation image indicator $LaTeX: \angle C$ $\angle C$ , the altitude, h, and side $LaTeX: \overline{AC}$ $\overline{AC}$ . We could write this trigonometric equation:

equation image indicator $LaTeX: \sin\left(85°\right)=\frac{h}{12}\\12\sin\left(85\right)=h$ $\sin\left(85°\right)=\frac{h}{12}\\12\sin\left(85\right)=h$

oblique triangle with sides of 12 in, 85°, and base of 8 in. with h=12sin(85)

So, now we have solved for the height of the triangle and can find the area of the triangle using our formula:

equation image indicator $LaTeX: A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(8\right)\left(12\sin85\right)\\ A\approx47.82in^2$ $A=\frac{1}{2}bh\\ A=\frac{1}{2}\left(8\right)\left(12\sin85\right)\\ A\approx47.82in^2$

Could we generalize this method to any triangle?

triangle ABC with sides a, b, and angle c

Say you know side a, b, and ∠C . If you draw an altitude from either unknown angle, you can use sine to solve for the height and then find the area.

triangle ABC with sides a, b, and angle c
sinC=b/a
h=a*sinC

equation image indicator $LaTeX: A=\frac{1}{2}bh\\ A=\frac{1}{2}b\left(a\:\sin C\right)\\ A=\frac{1}{2}ab\:\sin C$ $A=\frac{1}{2}bh\\ A=\frac{1}{2}b\left(a\:\sin C\right)\\ A=\frac{1}{2}ab\:\sin C$

This is an area formula for general triangles. What is important to remember when using this formula is that you need to know two sides and the included angle or the angle between those 2 sides.

Try these problems to see if you understand finding the area of triangles.

Heron's Formula

In the examples above, we were given the two side lengths and the angle between them. What if we were only given the three side lengths? The formula use to find the area of this triangle is called Heron's formula.

Heron's Formula.jpg

where $LaTeX: s=\frac{a+b+c}{2}$ $s=\frac{a+b+c}{2}$

Let's look at an example.

Example 1: Given the side lengths of the triangle, find the area using Heron's Formula.

We are given a = 19, b = 17, and c = 9. Use this to find s.

Since $LaTeX: s=\frac{a+b+c}{2}$ $s=\frac{a+b+c}{2}$ , substitute the values for a, b, and c to find s.

$LaTeX: s=\frac{19+17+9}{2}=\frac{45}{2}=22.5$ $s=\frac{19+17+9}{2}=\frac{45}{2}=22.5$

Now use Heron's formula to find the area.

$LaTeX: A=\sqrt{s(s-a)(s-b)(s-c)}$ $A=\sqrt{s(s-a)(s-b)(s-c)}$

$LaTeX: A=\sqrt{22.5(22.5-19)(22.5-17)(22.5-9)}=\sqrt{22.5(3.5)(5.5)(13.5)}=76.5units^2$ $A=\sqrt{22.5(22.5-19)(22.5-17)(22.5-9)}=\sqrt{22.5(3.5)(5.5)(13.5)}=76.5units^2$

Let's try another example.

Example 2: Find the area of the triangle, using Heron's formula.

Use the given side lengths to find s, first.

$LaTeX: s=\frac{a+b+c}{2}=\frac{12+13+14}{2}=\frac{39}{2}=19.5$ $s=\frac{a+b+c}{2}=\frac{12+13+14}{2}=\frac{39}{2}=19.5$

Now use Heron's formula to find the area.

$LaTeX: A=\sqrt{s(s-a)(s-b)(s-c)}$ $A=\sqrt{s(s-a)(s-b)(s-c)}$

$LaTeX: A=\sqrt{19.5(19.5-12)(19.5-13)(19.5-14)}=\sqrt{19.5(7.5)(6.5)(5.5)}=72.3units^2$ $A=\sqrt{19.5(19.5-12)(19.5-13)(19.5-14)}=\sqrt{19.5(7.5)(6.5)(5.5)}=72.3units^2$

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