AREA OF A TRIANGLE

AREA OF A TRIANGLE USING A BASE/HEIGHT PAIR

An earlier section, Area Formulas: Triangle, Parallelogram, Trapezoid,
showed how to find the area of a triangle with a base/height pair.
The key results are summarized here for convenience:

  • Take a vertex of a triangle.
    Drop a perpendicular to the opposite side or (if needed) to an extension of the opposite side.
  • The segment just created is called a height or altitude of the triangle. The side opposite the vertex is called the corresponding base.
  • Since every triangle has three vertices, every triangle has three base/height pairs.
  • Area is half base times height: $$ \begin{gather} \text{area of triangle}\cr \text{with base $\,b\,$ and height $\,h\,$}\cr = \frac 12 bh \end{gather} $$
base/height of triangle
Drop a perpendicular to the opposite side ...
base/height of triangle
... or, to an extension of the opposite side.


A derivation of the triangle area formula ($\,bh/2\,$) was given in
Area Formulas: Triangle, Parallelogram, Trapezoid for the situation
where the altitude ‘hits’ the base (left-most picture above).

The formula also works when the altitude doesn't ‘hit’ the base,
but instead hits an extension of the base, as shown at right.
Here's how:
  • Suppose you want the area of the blue triangle at right ($\,\triangle ABC\,$).
    This triangle has base $\,b\,$ and height $\,h\,$.
  • Let $\,x\,$ denote the length of the red-dashed segment.
Then: $$ \begin{align} &\text{area of blue triangle}\cr\cr &\qquad = \quad \text{area of big right triangle } (\triangle ABD) - \text{area of small right triangle } (\triangle CBD)\cr\cr &\qquad = \quad \frac 12(b+x)h - \frac 12 xh\cr\cr &\qquad = \quad \frac 12bh + \frac 12xh - \frac 12xh\cr\cr &\qquad = \quad \frac 12bh \end{align} $$

area of blue triangle is $\,\frac 12 bh\,$

Note that there are infinitely many triangles with a given base/height pair.
For example, all the triangles below have the same base/height pair.
Same base/height pair—same area!
A base/height pair uniquely determines the area of a triangle, but not the shape of the triangle.

All these triangles have the same base/height pair, hence the same area.


AREA OF A TRIANGLE USING TWO SIDES AND AN INCLUDED ANGLE

By the SAS (side-angle-side) triangle congruence theorem,
two sides and an included angle uniquely defines a triangle.

Let $\,a\,$ and $\,b\,$ denote two sides of a triangle,
with included angle $\,\theta\,$.

The sketches at right and below show the two situations that can occur when finding the altitude that has corresponding base $\,b\,$. In both cases, the altitude length is found using the yellow triangle (which is a right triangle with hypotenuse $\,a\,$).

In both cases, the altitude length is $\,h = a\sin\theta\,$.

Thus: $$ \begin{align} &\text{area of triangle with sides $a$ and $b$ and included angle $\theta$}\cr\cr &\qquad = \frac 12(\text{base})(\text{height})\cr\cr &\qquad = \frac 12 b(a\sin\theta)\cr\cr &\qquad = \frac 12 ab\sin\theta \end{align} $$ In words, the area is:

half the product of the two sides,
times the sine of the included angle



(When we write $\,180^\circ - \theta\,$, it is assumed that $\,\theta\,$ is measured in degrees.)

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Law of Sines
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