AMC 8 · 2011 · #20
Grade 8 geometry-2dProblem
Quadrilateral ABCD is a trapezoid, AD=15, AB=50, BC=20, and the altitude is 12. What is the area of the trapezoid?
Pick an answer.
AMC 8 2011 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: Trapezoid $ABCD$ has parallel sides $AB = 50$ (top) and $CD$ (bottom), slant legs $AD = 15$ and $BC = 20$, and a height of $12$ between the parallel sides. Find its area.
Givens: Top base $AB = 50$; Left leg $AD = 15$; Right leg $BC = 20$; Altitude (height between $AB$ and $CD$) $= 12$; Answer choices: (A) $600$, (B) $650$, (C) $700$, (D) $750$, (E) $800$
Unknowns: The area of trapezoid $ABCD$
Understand
Restated: Trapezoid $ABCD$ has parallel sides $AB = 50$ (top) and $CD$ (bottom), slant legs $AD = 15$ and $BC = 20$, and a height of $12$ between the parallel sides. Find its area.
Givens: Top base $AB = 50$; Left leg $AD = 15$; Right leg $BC = 20$; Altitude (height between $AB$ and $CD$) $= 12$; Answer choices: (A) $600$, (B) $650$, (C) $700$, (D) $750$, (E) $800$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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.
Execute — Answer: D
4.G.A.2 Step 1 - Drop a perpendicular from $A$ to $CD$, hitting it at $X$, and from $B$ to $CD$, hitting it at $Y$.
- Both new segments have length $12$ (the altitude).
- This carves the trapezoid into three pieces: left right triangle $ADX$, rectangle $ABYX$, and right right triangle $BCY$.
💡 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.
8.G.B.7 Step 2 - Use the Pythagorean theorem on the left right triangle $ADX$.
- Its legs are the altitude $AX = 12$ and the unknown horizontal piece $DX = a$; its hypotenuse is the slant leg $AD = 15$.
💡 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.7 Step 3 Repeat on the right right triangle $BCY$, with legs $BY = 12$ and $YC = b$ and hypotenuse $BC = 20$.
💡 Another familiar right triangle: $12, 16, 20$ is the $3, 4, 5$ family scaled by $4$.
4.G.A.2 Step 4 - Add the three horizontal pieces to get the bottom base.
- $CD = DX + XY + YC = a + 50 + b$.
💡 Reassembling the bottom edge from the rectangle's width plus the two triangle legs is the second half of the subproblem strategy.
6.G.A.1 Step 5 Apply the trapezoid area formula with $b_1 = AB = 50$, $b_2 = CD = 75$, and height $h = 12$.
💡 Knowing $\tfrac{1}{2}(b_1 + b_2)h$ — derivable by composing the trapezoid from triangles and a rectangle — is the Grade 6 "find area by decomposing" standard.
4.G.A.2 Drop a perpendicular from $A$ to $CD$, hitting it at $X$, and from $B$ to $CD$, 8.G.B.7 Use the Pythagorean theorem on the left right triangle $ADX$. Its legs are the a 8.G.B.7 Repeat on the right right triangle $BCY$, with legs $BY = 12$ and $YC = b$ and h 4.G.A.2 Add the three horizontal pieces to get the bottom base. $CD = DX + XY + YC = a + 6.G.A.1 Apply the trapezoid area formula with $b_1 = AB = 50$, $b_2 = CD = 75$, and heig Review
Reasonableness: Sanity check by adding the three pieces directly. Left triangle: $\tfrac{1}{2} \times 9 \times 12 = 54$. Middle rectangle: $50 \times 12 = 600$. Right triangle: $\tfrac{1}{2} \times 16 \times 12 = 96$. Total $= 54 + 600 + 96 = 750$, matching the formula answer (D). Magnitude check: the trapezoid is wider than $50$ and taller than $0$ but less than a $75 \times 12 = 900$ rectangle, and $750$ sits comfortably in that window.
Alternative: Tool #3 (Eliminate Possibilities) on the choices. The area must be $\tfrac{1}{2}(50 + CD)(12) = 6(50 + CD)$, so each choice fixes $CD$: (A) $600 \Rightarrow CD = 50$ (would make $ABCD$ a parallelogram with $a = b = 0$, impossible since $AD \ne BC$); (B) $650 \Rightarrow CD \approx 58.3$ (not an integer); (C) $700 \Rightarrow CD \approx 66.7$; (D) $750 \Rightarrow CD = 75$ ✓ (matches $9 + 50 + 16$); (E) $800 \Rightarrow CD \approx 83.3$. Only (D) gives integer leg pieces, the same conclusion.
CCSS standards used (min grade 8)
4.G.A.2Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines (Identifying $ABYX$ as a rectangle (so $XY = AB = 50$) after dropping the two altitudes, and naming the side pieces of the trapezoid.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Solving $a^2 + 12^2 = 15^2 \Rightarrow a = 9$ on the left triangle and $b^2 + 12^2 = 20^2 \Rightarrow b = 16$ on the right triangle.)6.G.A.1Find the area of right triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Computing the trapezoid area $\tfrac{1}{2}(50 + 75)(12) = 750$ — equivalently, summing the rectangle and two right triangles into which the trapezoid was decomposed.)
⭐ Drop 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 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.
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