Competition · AMC preparation · step 4 of 4
AMC 8 · 2007 · #8
Grade 6 geometry-2d
Pick an answer.
AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #1 (Draw a Diagram) is the lead because the answer pops out as soon as we annotate the figure: with AD ⊥ DC and BE ∥ AD, the piece ABED has two pairs of parallel sides and a right angle, so it is a rectangle — and since AD = AB = 3, it is a 3 × 3 square. Tool #7 (Identify Subproblems) then splits the trapezoid into the labeled square ABED plus the right triangle BEC, so the triangle's legs (BE and EC) read straight off the picture. No algebra needed.
Label the square
ABED has AB ∥ DE and AD ∥ BE, a right angle at D, and AD = AB = 3 — so it is a square of side 3.
A Grade 5 student classifies ABED by its properties: two pairs of parallel sides plus a right angle plus equal adjacent sides forces "square."
5.G.B.3Draw A DiagramRead the legs off the diagram
Since ABED is a square, BE = 3 and DE = 3, so EC = DC - DE = 6 - 3 = 3.
Splitting the trapezoid into "square piece" + "triangle piece" is the subproblem move. The triangle's base is whatever the square does not cover.
4.MD.A.3Identify SubproblemsFind the triangle's area
△ BEC is right-angled at E with legs BE = 3 and EC = 3, so its area is ½ · 3 · 3 = 4.5 → (B).
Once the legs are known, the Grade 6 area formula closes the problem in one line.
Triangle BEC is right-angled at E with legs BE = 3 and EC = 3, so its area is half the product of those two legs.
▸ Why?
The corner of △ BEC at E is a right angle, because BE stands perpendicular to the base line DC.
▸ Why?
BE is parallel to AD, and DC crosses both of these parallel segments; a crossing line makes matching angles equal, so the right angle that AD forms with DC at D is copied where BE meets DC at E.
▸ Why?
Each leg measures 3: the upright leg BE matches side AD, and the flat leg EC is what remains of DC after taking away DE.
▸ Why?
BE is the segment AD slid straight sideways to pass through B and E, and sliding a segment without turning it keeps its length, so BE = AD = 3.
▸ Why?
E lies on DC, cutting it into DE and EC; since DE is 3 and the two pieces together fill DC = 6, the leftover EC is 3.
▸ Why?
DE is side AB slid straight down onto the base line, and sliding keeps length, so DE = AB = 3.
▸ Why?
DE and EC meet at E with no gap and no overlap, so their lengths add up to the whole length DC.
▸ Why?
The area of this right triangle is half the product of its two legs, because the triangle is exactly one half of the rectangle built on those legs.
▸ Why?
Build the rectangle with sides BE and EC; its diagonal BC cuts it into two triangles that lay exactly onto each other, so △ BEC is one of two equal halves of that rectangle.
When a trapezoid has one slanted side, drop a perpendicular from the top corner. The trapezoid splits into a rectangle (or square) plus a right triangle, and the triangle's legs read straight off the picture.
- Label the square
- Read the legs off the diagram
- Find the triangle's area
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