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:

triangles

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

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1 Comment

  1. Area of a Triangle and Volume of Common Shapes: Index | Mean Green Math

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