AMC 10 · 2013 · #3
Grade 6 geometry-2dSquare ABCD has side length 10. Point E is on BC, and the area of △ABE is 40. What is BE?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A square with side length $10$ has a point $E$ somewhere on side $\overline{BC}$. The triangle formed by the two square-corners $A$ and $B$ together with $E$ has area $40$. Find the length $BE$.
Givens: Square $ABCD$ has side length $10$, so $AB = BC = 10$; $E$ lies on side $\overline{BC}$; Area of $\triangle ABE$ is $40$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Unknowns: The length $BE$
Understand
Restated: A square with side length $10$ has a point $E$ somewhere on side $\overline{BC}$. The triangle formed by the two square-corners $A$ and $B$ together with $E$ has area $40$. Find the length $BE$.
Givens: Square $ABCD$ has side length $10$, so $AB = BC = 10$; $E$ lies on side $\overline{BC}$; Area of $\triangle ABE$ is $40$; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $7$, (E) $8$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable
Tool #1 (Draw a Diagram) is primary because the whole problem turns on reading the figure correctly: the square's corner at $B$ is a right angle, and since $E$ sits on $\overline{BC}$, triangle $ABE$ is a right triangle whose two legs are the side $AB$ and the piece $BE$. Once the diagram shows those perpendicular legs, Tool #4 (Introduce a Variable) finishes the job: call $BE$ the unknown, write the right-triangle area as $\tfrac12 \cdot AB \cdot BE$, set it equal to $40$, and solve the one-step equation.
Execute — Answer: E
6.G.A.1 Step 1 Read the figure: a right triangle at B
- Look at where the triangle sits.
- The side $\overline{AB}$ runs up the left edge of the square and $\overline{BC}$ runs along the top; they meet at corner $B$, so they are perpendicular.
- Because $E$ is a point on $\overline{BC}$, the segment $\overline{BE}$ lies along that top edge.
- So triangle $ABE$ has a right angle at $B$, and its two legs are $AB = 10$ and the unknown $BE$.
💡 The corner of a square is a right angle, so the two square-edges meeting there are the perpendicular legs of the triangle.
6.G.A.1 Step 2 Write the area with BE as the base
- For a right triangle the two legs act as base and height.
- Take $BE$ as the base and $AB = 10$ as the height.
- Then the area is $\tfrac12 \cdot BE \cdot 10 = 5 \cdot BE$.
- Set this equal to the given area $40$.
💡 In a right triangle one leg is the base and the other leg is the height, so the area is just half their product.
6.EE.B.7 Step 3 Solve for BE
- The equation $5 \cdot BE = 40$ has the form $px = q$.
- Divide both sides by $5$ to get $BE = 8$.
- This is between $0$ and $10$, so $E$ really does land on the segment $\overline{BC}$.
- The answer is $\textbf{(E)}\ 8$.
💡 To undo a multiplication by $5$, divide both sides by $5$.
6.G.A.1 Look at where the triangle sits. The side $\overline{AB}$ runs up the left edge 6.G.A.1 For a right triangle the two legs act as base and height. Take $BE$ as the base 6.EE.B.7 The equation $5 \cdot BE = 40$ has the form $px = q$. Divide both sides by $5$ t Review
Reasonableness: Check the result against the picture. If $BE = 8$, the triangle's legs are $8$ and $10$, giving area $\tfrac12 \cdot 8 \cdot 10 = 40$, exactly as stated. Also $8$ is less than the side length $10$, so $E$ correctly stays on $\overline{BC}$ instead of past corner $C$. The full square has area $100$, and the triangle's $40$ is a sensible chunk of it, not larger than the whole.
Alternative: Work backwards from the area. A triangle of area $40$ with height $10$ must have base $\tfrac{2 \cdot 40}{10} = 8$; since that base is $BE$, we again get $BE = 8$. Or test the choices: only $BE = 8$ makes $5 \cdot BE$ equal $40$, so every other choice is eliminated.
CCSS standards used (min grade 6)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Recognizing $\triangle ABE$ as a right triangle with legs $AB=10$ and $BE$, and writing its area as $\tfrac12 \cdot BE \cdot 10$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Solving $5\,BE = 40$ to get $BE = 8$.)
⭐ At a square's corner the two edges are perpendicular, so the triangle is right-angled: its area is half of one leg times the other, and setting that equal to $40$ gives $BE = 8$.
⭐ At a square's corner the two edges are perpendicular, so the triangle is right-angled: its area is half of one leg times the other, and setting that equal to $40$ gives $BE = 8$.
More like this
Same archetype — closest grade level first.