Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #20
Grade 8 geometry-2d
Pick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure is given, but the key move — dropping altitudes from A and B down to CD — has to be added by hand. That is Tool #1 (Draw a Diagram): augment the picture so the structure is visible. Once those two altitudes are drawn, Tool #7 (Identify Subproblems) takes over: the trapezoid splits into a left right triangle (legs 12 and a, hypotenuse 15), a middle rectangle of width AB = 50, and a right right triangle (legs 12 and b, hypotenuse 20). Each piece is easy on its own, and CD = a + 50 + b falls out. No algebra (Tool #13) is needed.
Drop the two perpendiculars
Drop altitudes from A and B onto CD, splitting the trapezoid into rectangle ABYX of width 50 plus right triangles ADX and BCY.
Recognizing that ABYX is a rectangle (two pairs of parallel sides, all right angles) is the Grade 4 "classify two-dimensional figures" move, and it gives us XY = 50 for free.
4.G.A.2Draw A DiagramFind the left base piece
Pythagoras on right triangle ADX (legs 12 and a, hypotenuse AD = 15) gives horizontal piece a = 9.
This is the classic 9, 12, 15 right triangle (a 3, 4, 5 scaled by 3) — Tool #7 turns one trapezoid into a familiar right triangle.
8.G.B.7Identify SubproblemsFind the right base piece
Same on right triangle BCY (legs 12 and b, hypotenuse BC = 20) gives b = 16.
Another familiar right triangle: 12, 16, 20 is the 3, 4, 5 family scaled by 4.
8.G.B.7Identify SubproblemsAdd the three pieces
Add the three horizontal pieces to rebuild the bottom base: CD = 9 + 50 + 16 = 75.
Reassembling the bottom edge from the rectangle's width plus the two triangle legs is the second half of the subproblem strategy.
The bottom base CD of the trapezoid has length 75.
▸ Why?
The two altitudes meet CD at X and Y, cutting it into three pieces in a row — DX, then XY, then YC — so CD = DX + XY + YC = 9 + 50 + 16.
▸ Why?
X and Y sit on segment CD between D and C, splitting it into DX, XY, and YC with no gaps or overlaps, so the three lengths add back to the whole CD.
▸ Why?
The middle piece XY equals 50 because ABYX is a rectangle, and its opposite sides AB and XY have equal length.
▸ Why?
AX and BY are both altitudes of length 12, each perpendicular to AB and to CD, so sliding side AB straight down by 12 lays it exactly onto XY — a slide keeps every length, so XY = AB = 50.
▸ Why?
The left piece DX equals 9 because ADX is a right triangle whose two legs are the altitude AX = 12 and DX itself, and whose hypotenuse is the slant leg AD = 15, so the legs' squares add up to the hypotenuse's square: DX² + 12² = 15², giving DX² = 81 and DX = 9.
▸ Why?
The right piece YC equals 16 because BCY is a right triangle whose two legs are the altitude BY = 12 and YC itself, and whose hypotenuse is the slant leg BC = 20, so the legs' squares add up to the hypotenuse's square: YC² + 12² = 20², giving YC² = 256 and YC = 16.
Apply the trapezoid area formula
Plug bases 50 and 75 with height 12 into the trapezoid area formula to get area = 750.
Knowing 1/2(b₁ + b₂)h — derivable by composing the trapezoid from triangles and a rectangle — is the Grade 6 "find area by decomposing" standard.
6.G.A.1Identify SubproblemsDrop two altitudes and the trapezoid becomes a rectangle plus two right triangles — then Grade 8 Pythagorean theorem and the Grade 6 area formula finish it off.
- Drop the two perpendiculars
- Find the left base piece
- Find the right base piece
- Add the three pieces
- Apply the trapezoid area formula
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