AMC 10 · 2025 · #15

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglespythagorean-theoremsystems-of-equationsarea-triangles identify-subproblems ↑ Prerequisites: similar-trianglespythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram
Problem
In the figure, ABEF is a rectangle with AB=1 and AF=7. A segment AD of length 5 drops from A down to a point D, with AD ⊥ DE. The slanted side BE of the rectangle crosses AD at a point C. Find the area of △ ABC.

Pick an answer.

(A)
$\frac{3}{8}$
(B)
$\frac{4}{9}$
(C)
$\frac{1}{8}\sqrt{13}$
(D)
$\frac{7}{15}$
(E)
$\frac{1}{8}\sqrt{15}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The area of △ ABC needs the leg BC, and BC is hidden inside the crossing point C. The crossing creates two triangles, △ ABC and △ EDC, that share the vertical angle at C and each have a right angle, so they are similar. Tool #4 (Introduce a Variable) is the engine: name the two unknown pieces BC and AC, turn the similarity ratio and the two 'lengths add up' facts into a pair of linear equations, and solve. Tool #7 (Identify Subproblems) first splits off the smaller job of finding DE with the Pythagorean theorem, which fixes the similarity ratio. Tool #1 (Draw a Diagram) keeps track of which segment lies on which line so the collinear sums are written correctly.

1STEP 1

Read the rectangle and its diagonal

Opposite sides match, so BE=7. Draw diagonal AE: △ ABE is right-angled at B with legs 1 and 7, so Pythagoras gives AE.

AE=√(AB²+BE²)=√(1²+7²)=√(50)
2STEP 2

Find DE

△ ADE is right-angled at D, so with hypotenuse AE=√(50) and leg AD=5, Pythagoras leaves DE=5.

DE=√(AE²-AD²)=√(50-25)=√(25)=5
3STEP 3

Spot the similar triangles

At C, △ ABC and △ EDC each own a right angle (at B, at D) plus equal vertical angles, so AA similarity gives AB:ED=1:5.

△ ABC ∼ △ EDC, AB:ED=1:5
4STEP 4

Name the unknown pieces and build equations

Let BC=x and AC=y; the 1:5 ratio makes DC=5x and EC=5y, and the collinear sums AC+CD=5 and BC+CE=7 give two equations.

y+5x=5 and x+5y=7
5STEP 5

Solve the system

Substitute y=5-5x from the first equation into the second and solve: the leg we need is BC=3/4.

x+5(5-5x)=7 → x+25-25x=7 → -24x=-18 → x=3/4
6STEP 6

Compute the area

△ ABC is right-angled at B, so its legs AB=1 and BC=3/4 are base and height — the area is half their product.

[△ ABC]=1/2 · AB · BC=1/2 · 1·3/4=3/8 → (A)
Answer
3/8
Back-substitute x=3/4: then AC=y=5-5·3/4=5/4 and EC=5y=25/4. Check the BE length: BC+EC=3/4+25/4=7=BE, which matches. Also △ ABC should satisfy the Pythagorean theorem: AB²+BC²=1+9/16=25/16=(5/4)²=AC², consistent with AC=5/4. The area 3/8 is a small, clean fraction, fitting a tiny triangle wedged in the corner near B, and it is choice (A).
💡Key takeaway

When two lines cross, look for a pair of similar triangles from the equal vertical angles — name the mystery pieces, write two equations, and solve.

  • Read the rectangle and its diagonal
  • Find DE
  • Spot the similar triangles
  • Name the unknown pieces and build equations
  • Solve the system
  • Compute the area

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