Points $X$ and $Y$ lie on sides $AB$ and $AC$ respectively of triangle $ABC$. If $AB=7$, $AC=10$, $AX=4$ and $AY=6$, what is the ratio of the area of triangle $AXY$ to the area of triangle $ABC$?
However, I notice that they share the same angle at $A$, so I can use the formula for the area of a triangle in terms of the base and the angle. The area of triangle $ABC$ is $\frac{1}{2}AB \cdot AC \sin A$, and the area of triangle $AXY$ is $\frac{1}{2}AX \cdot AY \sin A$. To find the ratio of the areas, I can divide the second expression by the first one and simplify. The ratio is $\frac{\frac{1}{2}AX \cdot AY \sin A}{\frac{1}{2}AB \cdot AC \sin A} = \frac{AX \cdot AY}{AB \cdot AC}$. Now I can plug in the given values and calculate. The ratio is $\frac{4 \cdot 6}{7 \cdot 10} = \frac{24}{70} = \frac{12}{35}$.

