# SAT problem of the week, September 22, 2014 solution

Hint for process of elimination: In a ratio, the number before the word “to” belongs in the numerator, and the number after the word “to” belongs in the denominator. The SAT almost always give you ratios in the opposite order to test you, so you need to watch for that. Some of the choices are less than 1, which incorrectly imply that the area of $latex \triangle AED &s=1$ is smaller than the area of $latex \triangle ACB &s=1$. This eliminates choices B and C.

Also, note that unless the SAT says a figure is not drawn to scale, then the figure is almost definitely drawn to scale. That allows you to eliminate choices by estimating the size of the figures and eliminating choices that are too far off.

Read on for a full solution.

Since all the triangles are right triangles sharing the same $latex \angle A &s=1$, they are similar triangles, and therefore have proportional side lengths. Let $latex x &s=1$ be the length of base $latex \overline{AC} &s=1$ of $latex \triangle ACB &s=1$. Using the midpoint properties, the length of $latex \overline{CG} &s=1$ is also $latex x &s=1$, and the length of $latex \overline{CE} &s=1$ is $latex \dfrac{x}{2} &s=1$. Therefore the length of base $latex \overline{AE} &s=1$ of $latex \triangle AED &s=1$ is $latex \dfrac{3x}{2} &s=1$. Similarly, Let $latex y &s=1$ be the height of $latex \triangle ACB &s=1$. Then the height of $latex \triangle AED &s=1$ is $latex \dfrac{3y}{2} &s=1$. The ratio of the areas of the two triangles is then
$latex \dfrac{\dfrac{1}{2} \dfrac{3x}{2} \dfrac{3y}{2}}{\dfrac{1}{2}xy} = \dfrac{\left( \dfrac{3}{2} \right)^2 xy }{xy} = \dfrac{9}{4}&s=1$