Area of a Triangle

The area of a triangle is half the base times the height.

Formula for Area

The statement in the blue box gives us a way to get the area of any triangle. The 'base' of the triangle is any side you choose to make the base, i.e. whichever side you choose to have at the bottom of the diagram. The height is then the distance measured at right angles to that to the apex, the point of the triangle not on the base. The diagrams below show how the height and base are measured.

Notice particularly how the height and base are measured for the second triangle.

With Right-Triangles

Before we show why the formula works for all triangles we're going to look at the simpler case of right triangles.

For right triangles we can easily show that the formula is true provided the 'base' is one of the shorter sides. Because we can make a rectangle out of two copies of a right triangle the area of a right triangle is just half the area of the resulting rectangle, i.e half of the product of the lengths of the two shorter sides of the triangle. Here are some diagrams that show this:

Dividing other triangles into Right Triangles

Now that we have established how to calculate the area of a right triangle we can establish the formula for the area of any triangle by dividing it up into right triangles. Here we show that the triangle can be divided into two right triangles. We have 'h' is the height of the triangle, a and b are sides of the two smaller triangle, and together a+b is the length of the base.

The area of the two smaller triangles are ${\frac {ah}{2}}$ and ${\frac {bh}{2}}$ . The area of the whole triangle is the sum of these, i.e.

${\frac {ah}{2}}+{\frac {bh}{2}}={\frac {ah+bh}{2}}={\frac {(a+b)h}{2}}={\frac {(a+b)}{2}}h$

which is half the base times the height.

We do need to be careful however, since we could have a triangle with an obtuse angle. One way to deal with that is to allow the length a or b to be a negative number. However, to be on the safe side it is better to draw out a diagram for an obtuse angle triangle and check it separately.

In this diagram we now have that the length of the base is a-b and the area of the triangle is got by subtracting the area of the smallest right triangle from the largest right triangle.

${\frac {ah}{2}}-{\frac {bh}{2}}={\frac {ah-bh}{2}}={\frac {(a-b)h}{2}}={\frac {(a-b)}{2}}h$

Since the base has length a-b we again get that the area is half the base times the height.

The area of a triangle
is half the base times the height.

Heron's formula states that the area A of a triangle whose sides have lengths a, b, and c is

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

where $\displaystyle s$ is the semiperimeter of the triangle:

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

There is a proof here.

• Let us try this for the 3-4-5 triangle, which we know is a right triangle. We know its area. It's half that of the rectangle with sides 3x4. So the area $A$ is ${\frac {3\times 4}{2}}=6$ .

Let's see if Heron's formula works too.

$s={\frac {3+4+5}{2}}=6$

And using Heron's formula:

$A={\sqrt {s(s-a)(s-b)(s-c)}}={\sqrt {6(6-3)(6-4)(6-5)}}={\sqrt {6\cdot 3\cdot 2\cdot 1}}={\sqrt {36}}=6$

It worked.

Heron of Alexandria

The formula is believed to be due to Hero (or Heron) of Alexandria (10 – 70 AD), a Greek mathematician.

• Now over to you. We have an equilateral triangle with each side of length 1. The base of the triangle is 1, the height can be worked out by Pythagoras. It's

${\sqrt {1^{2}-{\frac {1}{2^{2}}}}}={\frac {\sqrt {3}}{2}}$

so the area $A$ of our equilateral triangle is ${\frac {\sqrt {3}}{4}}$ .

• Work out what the semi perimeter S is for this triangle, and put the values for S and the lengths of the sides into Heron's formula and compute the area, A.

$s=$

And using Heron's formula:

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

Resources

• Areas of Triangles, Wikibooks: Areas of Triangles
• Heron's Formula, Wikibooks: Trigonometry
• Area of a Triangle. Written notes created by Professor Esparza, UTSA.