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2013 AMC 10A Problem 3

2013 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 1 of 5 · about 1 min · position P1-10

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\triangle ABE$ is $40$ . What is $BE$ ?

A)
$4$
B)
$5$
C)
$6$
D)
$7$
E)
$8$

Answer

Key insight

Triangle ABE is right-angled at B with legs AB = 10 and BE, so its area is 5 times BE.

Solution

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 $\boxed{\textbf{(E)}\ 8}$ .

Why this works

When a triangle has a vertex at the corner of a square and one side along a square side, that square side is an altitude, so the area formula immediately turns into a one-variable equation. Look for right angles handed to you by squares and rectangles; they make "area given, find a length" problems one line long.

The trap

Forgetting the 1/2 in the triangle area formula and getting BE = 4.

Common mistakes

Forgetting the 1/2 in the triangle area formula and getting BE = 4.
Using $AE$ (the hypotenuse) as a side in the area formula.

Techniques

Set up the equation/formula and compute; no special trick needed