# Similarity and Scale Factor

If I have two triangles ABC and ADE and I know that each of the triangles is equilateral, each of the angles is 60 degrees. Is there a way to determine the scale factor of the two triangles, for example:

The ratio:

Area of ADE $:$Area of ABC

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If the ratio of corresponding lengths in two similar figures is $p:q$, then the ratio of their areas is $p^2:q^2$. (The ratio of their volumes would be $p^3:q^3$.) –  Blue Jun 5 '13 at 20:23
Yeah, I understand that Area of ADE $:$ Area of ABC $=$ $k^{2}$, where k is the scale factor. –  Sujaan Kunalan Jun 5 '13 at 20:26
Usually I am used to finding scale factor by dividing two sides but since no sides are given, I'm not sure how to find it. –  Sujaan Kunalan Jun 5 '13 at 20:27
Is there a figure? Without information beyond "each of the triangles is equilateral" (the $60^\circ$ thing is redundant) and that the triangles share vertex $A$, there's no way to determine the scale factor. Where are points $D$ and $E$? –  Blue Jun 5 '13 at 20:29
Give me a minute- I'll put up the figure. –  Sujaan Kunalan Jun 5 '13 at 20:34

If side $DC$$=$ $x$, then $AC$ must be $2x$ because both triangles are equilateral. By Pythagorean theorem, $AD$ is $3^\frac{1}{2}x$. Then $AD:AC=3^\frac{1}{2}:2=\frac{3^\frac{1}{2}{}}{2}:1$. So the scale factor $k$, is $\frac{3^\frac{1}{2}}{2}$. Using the formula:
$\frac{\text{Area of }AED}{\text{Area of }ABC}=k^2$
$\frac{\text{Area of }AED}{\text{Area of }ABC}=(\frac{3^\frac{1}{2}}{2})^2$
$\frac{\text{Area of }AED}{\text{Area of }ABC}=\frac{3}{4}$
Therefore $k^2= \frac{3}{4}$ and so the ratio between triangle $ADE$ and $ABC$ is $1:\frac{3}{4}$
(+1) I'd have written the answer as $4:3$, personally, but that's just a matter of taste. –  Cameron Buie Jun 5 '13 at 23:20