2013 AMC 10A Problem 3

P1-10~1 minPrint

Triangles: Area & Pythagorean Set up the equation/formula and compute; no special trick needed

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

Show solution without answering

Key insight

The corner of the square at B is a right angle, so AB and BE are the base and height of the triangle; half of 10 times BE equals 40.

Solution

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Geometry, Chapter 10: Geometric Inequalities (§10.3)