AMC 10 · 2021 · #21
Grade 8 geometry-2d
Pick an answer.
Tool #7 (Identify Subproblems) opens the problem: the diagonal AC cuts the trapezoid into △ ABC and △ ACD, and the two perpendiculars we were handed are exactly the heights of those triangles onto the common base AC. So the area is 1/2(AC)(BY) + 1/2(AC)(DX), and everything reduces to two heights. Tool #1 (Draw a Diagram) supplies the second fact: the parallel sides make the two right triangles CYB and AXD similar, which locks the two heights into a fixed ratio. Tool #4 (Introduce a Variable) names the heights and then swaps in the single variable u for a squared height, turning a messy equation into a linear one. Tool #13 (Convert to Algebra) turns AB = CD into an equation via the Pythagorean theorem. Tool #16 (Change Focus) is the finishing move: the area only needs the sum of the two heights, so square the sum instead of digging out each height separately — the ugly radicals never appear.
Cut the trapezoid along its diagonal
Cut it into two pieces along the diagonal.
The two perpendiculars in the picture are not decoration — they are the heights of the two halves the diagonal creates, so the whole area is just 3 times their sum.
6.G.A.1Identify SubproblemsParallel sides make similar right triangles
The parallel sides give similar triangles.
Parallel sides mean the diagonal leans away from each of them at the same angle, so the two right triangles hanging off the diagonal are the same shape at different sizes.
Parallel sides mean the diagonal leans away from each of them at the same angle, so the two right triangles match in shape.
▸ Why?
A line crossing two parallels makes matching angles at both crossings.
▸ Why?
Triangles with the same angles have all their matching sides in one fixed ratio.
Pythagoras on each leg of the trapezoid
Apply Pythagoras to each leg.
Each perpendicular splits a leg of the trapezoid off as the hypotenuse of a right triangle whose other two sides you already know.
8.G.B.7Convert To AlgebraCash in the isosceles condition
The isosceles condition gives an equation.
"The two legs are equal" is the last unused fact, and it is exactly the one equation needed to pin down the size of the picture.
8.EE.C.7Introduce A VariableChase the sum, not the pieces
Chase the sum, not the pieces.
Asking for the sum instead of each height keeps the arithmetic on squares, where the numbers stay whole.
8.EE.A.2Change Focus Count The ComplementAssemble the area
Assembling gives three root thirty-five.
The whole chase was only ever about one number, p + q, and multiplying it by 3 finishes the job.
7.G.B.6Identify SubproblemsCut the trapezoid along its diagonal and the two given perpendiculars become the heights of the two halves, so the area is just 3(p+q); the parallel sides force p = 3/2q, the equal legs force q² = 28/5, and squaring the sum gives (p+q)² = 35 — area 3√(35), no messy roots needed.
- Cut the trapezoid along its diagonal
- Parallel sides make similar right triangles
- Pythagoras on each leg of the trapezoid
- Cash in the isosceles condition
- Chase the sum, not the pieces
- Assemble the area
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