# Area of a triangle: Base and height (Part 1)

This begins a series of post concerning how the area of a triangle can be computed. This post concerns the formula that students most often remember:

$A = \displaystyle \frac{1}{2} b h$

Why is this formula true? Consider $\triangle ABC$, and form the altitude from $B$ to line $AB$. Suppose that the length of $AC$ is $b$ and that the altitude has length $h$. Then one of three things could happen:

Case 1. The altitude intersects $AC$ at either $A$ or $C$. Then $\triangle ABC$ is a right triangle, which is half of a rectangle. Since the area of a rectangle is $bh$, the area of the triangle must be $\displaystyle \frac{1}{2} bh$.

Knowing the area of a right triangle will be important for Cases 2 and 3, as we will act like a good MIT freshman and use this previous work.

Case 2. The altitude intersects $AC$ at a point $D$ between $A$ and $C$. Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle ABD + \hbox{~Area of~} \triangle BCD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} b_1 h + \frac{1}{2} b_2 h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b_1 + b_2) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$

Case 3. The altitude intersects $AC$ at a point $D$ that is not in between $A$ and $C$. Without loss of generality, suppose that $A$ is between $D$ and $C$. Then $\triangle ABD$ and $\triangle BCD$ are right triangles, and so

$\hbox{Area of~} \triangle ABC = \hbox{Area of ~} \triangle BCD - \hbox{~Area of~} \triangle ACD$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t) h + \frac{1}{2} t h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} (b+t-t) h$

$\hbox{Area of~} \triangle ABC = \displaystyle \frac{1}{2} bh$