2005 AMC 10A Problem 25

Both triangles $ADE$ and $ABC$ start at corner $A$. Grow the small one into the big one in two steps, changing one side at a time.

1. Draw $BE$. This gives a middle triangle $ABE$ to compare with.
2. From $ADE$ to $ABE$. Both triangles have their top at $E$ and their bases on line $AB$, so they have the same height from $E$. Their areas compare as their bases:
$$\frac{\text{area of } ADE}{\text{area of } ABE} = \frac{AD}{AB} = \frac{19}{25}.$$
3. From $ABE$ to $ABC$. Both have their top at $B$ and their bases on line $AC$, so again the same height:
$$\frac{\text{area of } ABE}{\text{area of } ABC} = \frac{AE}{AC} = \frac{14}{42} = \frac{1}{3}.$$
4. Combine. Multiply the two ratios:
$$\frac{\text{area of } ADE}{\text{area of } ABC} = \frac{19}{25} \times \frac{1}{3} = \frac{19}{75}.$$
5. Find the quadrilateral. Think of triangle $ABC$ as $75$ equal pieces of area. Triangle $ADE$ is $19$ of them. The quadrilateral $BCED$ is everything else: $75 - 19 = 56$ pieces.
6. Answer the question asked. The ratio of $ADE$ to $BCED$ is $\frac{19}{56}$.

The answer is $\boxed{\textbf{(D)}\ \frac{19}{56}}$.

Two triangles that share an angle have areas in the same ratio as the products of the two sides around that angle. Here that is $19 \times 14$ against $25 \times 42$. The length $BC = 39$ is never used; it is there to tempt you into a long calculation. Finally, read the question carefully: it wants the small triangle compared with what is left, not with the whole.

Use the area formula with an angle: the area of a triangle is $\frac{1}{2} \times$ (two sides) $\times$ (sine of the angle between them). Both triangles use the same angle at $A$, so the sine cancels:
$$\frac{\text{area of } ADE}{\text{area of } ABC} = \frac{19 \times 14}{25 \times 42} = \frac{19}{75},$$
and the part that is left is $75 - 19 = 56$, giving $\frac{19}{56}$.