AMC 10 · 2004 · #9

Grade 6 geometry-2d
Archetype Overlap / Complement Counting └ Sub-archetype Geometric Area via Inclusion-Exclusion or Area Balance
area-trianglesarea-difference convert-to-algebraidentify-subproblems ↑ Prerequisites: area-triangles
📏 Medium solution 💡 1 insight 📊 Diagram
Problem
Two triangles △ ABE and △ ABC share the side AB, with a right angle at A (between AE and AB) and a right angle at B (between AB and BC). Given AB=4, BC=6, and AE=8, the segments AC and BE cross at D. Find [△ ADE] - [△ BDC], the difference of those two areas.

Pick an answer.

(A)
$\ 2$
(B)
$\ 4$
(C)
$\ 5$
(D)
$\ 8$
(E)
$\ 9$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The two triangles we are asked about, △ ADE and △ BDC, are the awkward slivers left after the diagonals cross — finding either one directly would need the location of D. The unlock is to stop chasing the slivers and look at the whole triangles they belong to. Point D sits on BE, so △ ADE is just △ ABE with the shared corner △ ABD removed; likewise △ BDC is △ ABC with that same △ ABD removed. That is a classic change of focus (Tool #16): add the common overlap back to both pieces instead of computing the pieces. The two full triangles are easy right triangles, so their areas come straight from the legs (Tool #1 to see which segments are the legs). Naming the shared overlap with a single variable (Tool #4) makes the cancellation visible: the unknown region drops out when we subtract, so we never need to find D at all.

1STEP 1

Area of each whole right triangle

Both triangles sharing AB are right, so area is half the leg product: [△ ABE] = 1/2·4·8 = 16 and [△ ABC] = 1/2·4·6 = 12.

[△ ABE] = 1/2·4·8 = 16, [△ ABC] = 1/2·4·6 = 12
2STEP 2

Name the shared overlap

The two triangles overlap in the corner region △ ABD where the diagonals cross; we never need its area, so let x = [△ ABD].

let x = [△ ABD]
3STEP 3

Write each target triangle as whole minus overlap

D on BE splits △ ABE, so [△ ADE] = 16 - x; D on AC splits △ ABC, so [△ BDC] = 12 - x.

[△ ADE] = 16 - x, [△ BDC] = 12 - x
4STEP 4

Subtract and watch the unknown cancel

Subtract: (16 - x) - (12 - x) = 16 - 12 = 4, choice (B). The same -x cancels, so where D sits never mattered.

[△ ADE] - [△ BDC] = (16 - x) - (12 - x) = 4 → (B)
Answer
4
The result equals [△ ABE] - [△ ABC] = 16 - 12 = 4, which makes sense: subtracting the two whole triangles removes their shared region automatically, so the sliver difference must match the whole-triangle difference. The number is small and positive, consistent with △ ADE sitting inside the taller triangle and △ BDC inside the shorter one. It matches choice (B), and note the trap answers: 8 (D) is the leg difference AE-BC=8-6, and 2 (A) is BC-AB — both are tempting but ignore the clean cancellation.
💡Key takeaway

When two shapes overlap in the same messy piece, don't fight to measure that piece — give it a name and compare the whole shapes instead. The shared part cancels, and here the answer is just 16 - 12 = 4.

  • Area of each whole right triangle
  • Name the shared overlap
  • Write each target triangle as whole minus overlap
  • Subtract and watch the unknown cancel

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