AMC 10 · 2004 · #9
Grade 6 geometry-2dIn the overlapping triangles △ABC and △ABE sharing common side AB, ∠EAB and ∠ABC are right angles, AB=4, BC=6, AE=8, and AC and BE intersect at D. What is the difference between the areas of △ADE and △BDC?
Pick an answer.
AMC 10 2004 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: Two triangles $\triangle ABE$ and $\triangle ABC$ 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 $\overline{AC}$ and $\overline{BE}$ cross at $D$. Find $[\triangle ADE] - [\triangle BDC]$, the difference of those two areas.
Givens: $\triangle ABE$ and $\triangle ABC$ share the common side $AB$, with $AB = 4$.; $\angle EAB = 90^\circ$, so $AE \perp AB$, and $AE = 8$.; $\angle ABC = 90^\circ$, so $BC \perp AB$, and $BC = 6$.; $E$ and $C$ lie on the same side of line $AB$, and diagonals $\overline{AC}$ and $\overline{BE}$ meet at $D$.; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $8$, (E) $9$
Unknowns: The difference between the area of $\triangle ADE$ and the area of $\triangle BDC$.
Understand
Restated: Two triangles $\triangle ABE$ and $\triangle ABC$ 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 $\overline{AC}$ and $\overline{BE}$ cross at $D$. Find $[\triangle ADE] - [\triangle BDC]$, the difference of those two areas.
Givens: $\triangle ABE$ and $\triangle ABC$ share the common side $AB$, with $AB = 4$.; $\angle EAB = 90^\circ$, so $AE \perp AB$, and $AE = 8$.; $\angle ABC = 90^\circ$, so $BC \perp AB$, and $BC = 6$.; $E$ and $C$ lie on the same side of line $AB$, and diagonals $\overline{AC}$ and $\overline{BE}$ meet at $D$.; Answer choices: (A) $2$, (B) $4$, (C) $5$, (D) $8$, (E) $9$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The two triangles we are asked about, $\triangle ADE$ and $\triangle BDC$, 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 $\overline{BE}$, so $\triangle ADE$ is just $\triangle ABE$ with the shared corner $\triangle ABD$ removed; likewise $\triangle BDC$ is $\triangle ABC$ with that same $\triangle ABD$ 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.
Execute — Answer: B
6.G.A.1 Step 1 Area of each whole right triangle
- Both triangles that share $AB$ are right triangles, so each area is half the product of its two legs.
- In $\triangle ABE$ the right angle at $A$ makes $AB=4$ and $AE=8$ the legs, giving area $\tfrac12\cdot4\cdot8 = 16$.
- In $\triangle ABC$ the right angle at $B$ makes $AB=4$ and $BC=6$ the legs, giving area $\tfrac12\cdot4\cdot6 = 12$.
💡 A right angle hands you the base and height for free — the two legs are perpendicular, so no extra work is needed to find a height.
6.EE.B.6 Step 2 Name the shared overlap
- The two big triangles are not separate: they both cover the corner region $\triangle ABD$ where the diagonals cross.
- We do not know its area and we will not need to, so let $x$ stand for it: $x = [\triangle ABD]$.
- Giving the messy shared piece a single name lets us track it through the algebra.
💡 When a quantity is hard to find but appears in both things you are comparing, name it and let it cancel later rather than fighting to compute it.
6.G.A.1 Step 3 Write each target triangle as whole minus overlap
- Since $D$ lies on $\overline{BE}$, segment $\overline{AD}$ cuts $\triangle ABE$ into exactly two pieces, $\triangle ABD$ and $\triangle ADE$, so $[\triangle ADE] = 16 - x$.
- Since $D$ lies on $\overline{AC}$, segment $\overline{BD}$ cuts $\triangle ABC$ into $\triangle ABD$ and $\triangle BDC$, so $[\triangle BDC] = 12 - x$.
- Each sliver is its whole triangle with the shared corner removed.
💡 Adding the same overlap to each sliver rebuilds a full, easy triangle — so the sliver is just the full area minus that shared overlap.
6.EE.A.3 Step 4 Subtract and watch the unknown cancel
- Take the difference the problem asks for.
- Both expressions carry the same $-x$, so the unknown overlap cancels: $(16 - x) - (12 - x) = 16 - 12 = 4$.
- The answer does not depend on where $D$ is, which is why we never had to locate it.
- The difference of the areas is $4$, choice (B).
💡 Subtracting two amounts that share the same unknown makes it vanish, leaving only the difference of the parts you actually know.
6.G.A.1 Both triangles that share $AB$ are right triangles, so each area is half the pro 6.EE.B.6 The two big triangles are not separate: they both cover the corner region $\tria 6.G.A.1 Since $D$ lies on $\overline{BE}$, segment $\overline{AD}$ cuts $\triangle ABE$ 6.EE.A.3 Take the difference the problem asks for. Both expressions carry the same $-x$, Review
Reasonableness: The result equals $[\triangle ABE] - [\triangle ABC] = 16 - 12 = 4$, 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 $\triangle ADE$ sitting inside the taller triangle and $\triangle BDC$ 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.
Alternative: Place coordinates $A=(0,0)$, $B=(4,0)$, $C=(4,6)$, $E=(0,8)$. Line $AC$ is $y=\tfrac32 x$ and line $BE$ is $y=-2x+8$; setting them equal gives $D=\left(\tfrac{16}{7}, \tfrac{24}{7}\right)$. Using the shoelace formula, $[\triangle ADE]=\tfrac{40}{7}$ and $[\triangle BDC]=\tfrac{12}{7}$, whose difference is $\tfrac{28}{7}=4$ — the same answer, but with far more arithmetic, confirming the elegance of the cancellation.
CCSS standards used (min grade 6)
6.G.A.1Find the area of triangles and other polygons by composing into rectangles or decomposing into triangles and other shapes (Getting each right triangle's area from its two legs, and decomposing each whole triangle into the shared corner plus the target sliver.)6.EE.B.6Use variables to represent numbers and write expressions when solving a problem (Naming the unknown shared overlap $\triangle ABD$ as $x$ so it can be carried symbolically through the comparison.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Simplifying $(16 - x) - (12 - x)$ so the unknown $x$ cancels and the difference reduces to $4$.)
⭐ 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 $16 - 12 = 4$.
⭐ 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 $16 - 12 = 4$.
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