8.7
Ratios of Areas
We conclude our consideration of area by looking at the comparisons of areas. There are two primary kinds of issues. Similar figures have areas that can be compared one way. Figures that share a common part have areas that are compared in a different way.
We start with an underlying principle concerning similar figures. When two figures are similar, the ratio of similarity of corresponding lengths is the same, no matter what parts of the similar figures those might be. We know that corresponding sides of similar polygons are in the same ratio; that's part of the definition of what makes figures similar to each other. And we have also shown that corresponding altitudes of similar triangles are in the same ratio as the corresponding sides — as are corresponding angle bisectors and corresponding medians. But it goes farther than that. Given two similar figures, if you select any two pairs of corresponding lengths (including perimeters), their ratios will be the same.
- Theorem 8.10:
The ratio of the areas of two similar triangles is the square of the
ratio of corresponding lengths.
Proof:
Consider two triangles $\triangle ABC$ and $\triangle XYZ$, with
$\triangle ABC\sim \triangle XYZ$. Let the ratio of similarity be
$k$. Let $\overline{CD}$ be the altitude to side $\overline{AB}$
in $\triangle ABC$; let $\overline{ZW}$ be the altitude to side
$\overline{XY}$ in $\triangle XYZ$. Then $\frac{AB}{XY}=k$ and
$\frac{CD}{ZW}=k$. When we consider the ratio of the areas of the
triangles, we have \begin{align} \frac{a(\triangle ABC)}{a(\triangle XYZ)}&=\frac{\frac{1}{2}(AB)(CD)}{\frac{1}{2}(XY)(ZW)}\\ &=\left(\frac{AB}{XY}\right)\left(\frac{CD}{ZW}\right)\\ &=k\cdot k\\ &=k^2. \end{align} |
In computing areas, we always multiply two lengths together. For a rectangle, it is length times width; for a square, it is side times side; for a triangle, it is base time height (times one-half); for a circle, it is radius times radius (times $\pi$); and so on. As a result, the proof above can be mimicked with any two similar figures, and we will always have the ratio of the areas of two similar figures as the square of the ratio of corresponding parts.
Example: |
In the figure,
$\overline{DE}\parallel\overline{AB}$, Determine $\frac{a(\triangle
DEC)}{a(ABED)}$. |
|
Solution: |
$\triangle DEC\sim
\triangle ABC$, with a ratio similarity $\frac{5}{8}$. Hence,
$$\frac{a(\triangle DEC)}{a(\triangle
ABC)}=\left(\frac{5}{8}\right)^2=\frac{25}{64}.$$ That means that
$\triangle DEC$ is $\frac{25}{64}$ of the whole triangle ($\triangle
ABC$). In that case, trapezoid, $ABED$ must be the remaining
$\frac{39}{64}$ (found by subtracting $\frac{25}{64}$ from 1) of the
whole thing. Then $$\frac{a(\triangle
DEC)}{a(ABED)}=\frac{\frac{25}{64}}{\frac{39}{64}}.$$ |
We now
come to a second means of comparing areas. When we compute the
ratio of the areas of $\triangle ABC$ and $\triangle XYZ$, we have
$$\frac{a(\triangle ABC)}{a(\triangle
XYZ)}=\frac{\frac{1}{2}(AB)(CD)}{\frac{1}{2}(XY)(ZW)},$$ where
$\overline{CD}$ and $\overline{ZQ}$ are the altitudes to
$\overline{AB}$ and $\overline{XY}$, respectively. If the
triangles
have equal bases but unequal altitudes, then the ratio of their areas
will simply be the ratio of the corresponding altitudes. In the
same
way, if they have equal altitudes but unequal bases, then the ratio of
their areas will be the ratio of the bases. And if the triangles
have
equal bases and equal altitudes, then even though they may not be
congruent, they will have the same area. |
Example: |
In trapezoid $ABCD$ shown to the
right, $\overline{AB}\parallel\overline{CD}$. What is the ration
of the area of $\triangle ABC$ to the area of $\triangle ADC$? |
|
Solution: |
Since
$\overline{AB}\parallel\overline{CD}$, the altitude to $\overline{AB}$
drawn from $C$ and the altitude drawn to $\overline{CD}$ from $A$ have
the same length. Hence the ration of the areas of the triangles
is the ration of the bases: $\frac{8}{5}$. |
A more
particular case of the situation in the example occurs in the figure to
the right. Lines $l$ and $m$ are parallel. That means that
in $\triangle ABC$ and $\triangle ABD$, the altitudes drawn to
$\overline{AB}$ are equal. Hence the triangles have the same
area. |