2026 AMC 10A Problem 3

Triangle $BEF$ has side $\overline{BE}$ lying along the bottom of the square, so use $\overline{BE}$ as the base. Its length is easy: $E$ is the midpoint of a side of length $24$, so $BE = 12$. The only unknown is the height, which is how far $F$ sits above the bottom side. Coordinates find that directly.

1. Place the square on a grid. Put $B$ at the origin, as in the picture:
$$B = (0, 0),\quad C = (24, 0),\quad D = (24, 24),\quad A = (0, 24).$$
The midpoint of $\overline{BC}$ is $E = (12, 0)$.

2. Line $BD$. It goes from $(0,0)$ to $(24,24)$, rising $1$ for every $1$ across. Its equation is
$$y = x.$$

3. Line $AE$. It starts at $(0, 24)$ and ends at $(12, 0)$. It drops $24$ while moving $12$ across, so its slope is $-24 \div 12 = -2$. Its equation is
$$y = 24 - 2x.$$
(Check: at $x = 12$ this gives $y = 0$, which is point $E$.)

4. Where they cross. At $F$ both equations hold, so
$$x = 24 - 2x \quad\Longrightarrow\quad 3x = 24 \quad\Longrightarrow\quad x = 8,$$
and then $y = x = 8$. So $F = (8, 8)$.

5. The area. The base $\overline{BE}$ lies on the line $y = 0$ and has length $12$. The height is the distance from $F$ up to that line, which is the $y$-coordinate of $F$, namely $8$. So
$$\text{area} = \tfrac12 \times 12 \times 8 = 48.$$

The answer is $\boxed{\textbf{(D)}\ 48}$.

When a triangle has one side lying on a side of a square (or on any line you know well), the area comes down to one number: how far the third corner is from that line. Two lines crossing inside a square are easy to write as equations once the square sits on a grid with a corner at the origin, and the crossing point gives that height immediately.

Use similar triangles instead of coordinates.

1. Sides $\overline{BC}$ and $\overline{AD}$ of the square are parallel. Look at triangle $FBE$ (at the bottom) and triangle $FDA$ (at the top).
2. The angles at $F$ are equal because they are vertical angles. The angles at $B$ and at $D$ are equal because $\overline{BD}$ crosses the two parallel sides, making alternate angles. So the two triangles are similar.
3. Their matching sides $BE = 12$ and $DA = 24$ are in the ratio $1 : 2$. So the whole bottom triangle is half the size of the top triangle, and in particular its height (from $F$ down to $\overline{BC}$) is half the top triangle's height (from $F$ up to $\overline{AD}$).
4. These two heights add up to the side of the square, $24$. Splitting $24$ in the ratio $1 : 2$ gives $8$ and $16$. So $F$ is $8$ above $\overline{BC}$.
5. The area of triangle $BEF$ is $\tfrac12 \times 12 \times 8 = 48$.