2013 AMC 10A Problem 3

1. Find a right angle. Every corner of a square is a right angle, so the angle at $B$ is $90^\circ$. The point $E$ is on side $BC$, so the angle $ABE$ is also $90^\circ$.
2. Pick the base and height. In a right triangle the two sides that meet at the right angle work as base and height. Here they are $AB = 10$ and $BE$.
3. Use the area. Area of a triangle is $\tfrac12 \times \text{base} \times \text{height}$:
$$\frac12 \times 10 \times BE = 40.$$
4. Solve. Half of $10$ is $5$, so $5 \times BE = 40$, and $BE = 8$.
5. Check it makes sense. $E$ must be on side $BC$, which has length $10$, and $8$ is less than $10$. Good.

The answer is $\boxed{\textbf{(E)}\ 8}$.

Squares and rectangles hand you right angles for free. When a triangle uses a corner of the square, the two sides at that corner are already a base and a height. So "the area is given, find a length" becomes one short equation. Always remember the $\tfrac12$ in the triangle area formula.

Since $ABCD$ is a square, $\angle ABC = 90^\circ$, and $E$ lies on side $BC$, so $\angle ABE$ is also a right angle. Triangle $ABE$ is therefore a right triangle with legs $AB = 10$ and $BE$.

Its area is half the product of the legs:
$$\frac12 \cdot 10 \cdot BE = 40 \quad\Longrightarrow\quad 5 \cdot BE = 40 \quad\Longrightarrow\quad BE = 8.$$

This is consistent with $E$ being on $BC$, since $8 \le 10$. The answer is $\boxed{\textbf{(E)}\ 8}$.