Formula for the ratio of areas of two similar triangles

geometrytriangles

Imagine we have two similar triangles ABC and DEF, and we need to find the ratio between the areas.

The formula for the area is $\frac{1}{2}hb$, where the $h$ is the height and $b$ is the base.

So in our case where we have bases $b$ and $e$ the ratio of the areas would be $\frac{\frac{1}{2}h_1b}{\frac{1}{2}h_2e}$

I see this formula shortened to just the ratio of their bases squared, so $\frac{b^2}{e^2}$.

First question: do we always put the smaller triangle in the numerator and the larger in the denominator?

Second question: why can the formula for the ratio between areas be shortened like that, is it because in the ratio between $h_1$ and $h_2$ is the same as $b$ and $e$ so we just swap the heights for bases?

Best Answer

By definition of similar triangles if Triangle 1 has Height $h_1$ and base $b_1$ and Triangle 2 has height $h_2$ and base $b_2$, then $h_2=kh_1$, $b_2=kb_1$ for some constant $k$.

$k=\frac{b_2}{b_1}=\frac{h_2}{h_1}$

Area 1 is $A_1=(1/2) h_1b_1$.

$A_2=(1/2)h_2b_2$

$\frac{A_2}{A_1}=(\frac{h_2}{h_1})(\frac{b_2}{b_1})=k^2=(\frac{b_2}{b_1})^2=(\frac{h_2}{h_1})^2$

Generally speaking, the ratio of the areas between two objects is the square of the ratio of their corresponding lengths.

If you consider $\frac{A_1}{A_2}=\frac{1}{k^2}$ you find the ratio between areas changes, but the principle still holds since $\frac{1}{k^2}=(\frac{b_1}{b_2})^2$

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