AMC 10 · 2002 · #25
Grade 8 geometry-2dIn trapezoid ABCD with bases AB and CD, we have AB=52, BC=12, CD=39, and DA=5. The area of ABCD is
Pick an answer.
AMC 10 2002 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: A trapezoid $ABCD$ has parallel sides (bases) $AB=52$ and $CD=39$, and slanted sides (legs) $BC=12$ and $DA=5$. Find the area enclosed by the trapezoid.
Givens: The two bases are parallel: $AB=52$ and $CD=39$; The two legs are $BC=12$ and $DA=5$; Area of a trapezoid $=\tfrac{1}{2}(\text{base}_1+\text{base}_2)\times\text{height}$; Answer choices: (A) $182$, (B) $195$, (C) $210$, (D) $234$, (E) $260$
Unknowns: The height (perpendicular distance) between the two bases; The area of trapezoid $ABCD$
Understand
Restated: A trapezoid $ABCD$ has parallel sides (bases) $AB=52$ and $CD=39$, and slanted sides (legs) $BC=12$ and $DA=5$. Find the area enclosed by the trapezoid.
Givens: The two bases are parallel: $AB=52$ and $CD=39$; The two legs are $BC=12$ and $DA=5$; Area of a trapezoid $=\tfrac{1}{2}(\text{base}_1+\text{base}_2)\times\text{height}$; Answer choices: (A) $182$, (B) $195$, (C) $210$, (D) $234$, (E) $260$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The height is the missing piece — once we have it, the area formula finishes the job. To get the height, drop a perpendicular from each top corner ($C$ and $D$) straight down to the bottom base $AB$. This splits the trapezoid into a middle rectangle (width $39$, the same as $CD$) plus two right triangles, one at each end. The legs $5$ and $12$ become the slanted sides (hypotenuses) of those triangles, and the two horizontal feet together fill the leftover $52-39=13$. Naming the left foot $x$ and the common height $h$ (tool #4) turns each right triangle into a Pythagorean equation. Two equations, two unknowns — solve for $h$, then plug into the area formula.
Execute — Answer: C
6.G.A.1 Step 1 Drop perpendiculars to make right triangles
- Drop a straight vertical line from $D$ and from $C$ down to the bottom base $AB$, landing at points $P$ and $Q$.
- The middle piece $PQCD$ is a rectangle, so $PQ=CD=39$.
- That leaves the two end pieces $AP$ and $QB$ to share the extra length $AB-PQ=52-39=13$.
- Each end piece is a right triangle: the left one has hypotenuse $DA=5$, the right one has hypotenuse $BC=12$, and both have the same vertical height $h$ (the height of the trapezoid).
💡 Cutting straight down from the short base turns the slanted trapezoid into a plain rectangle plus two right triangles you can measure.
8.G.B.7 Step 2 Name the parts and write Pythagoras
- Let the left horizontal foot $AP=x$; then the right foot $QB=13-x$.
- Both triangles rise the same height $h$.
- Apply the Pythagorean theorem to each right triangle: for the left triangle the legs $x$ and $h$ meet the hypotenuse $5$, and for the right triangle the legs $13-x$ and $h$ meet the hypotenuse $12$.
💡 Each slanted leg is the hypotenuse of a right triangle, so its length squared equals its horizontal run squared plus the height squared.
8.EE.C.7 Step 3 Subtract to find the horizontal foot
- Both equations contain $h^2$, so subtract the first from the second to make $h^2$ disappear.
- The $x^2$ terms also cancel, leaving a plain linear equation in $x$: $(13-x)^2-x^2=144-25$.
- Expand $(13-x)^2=169-26x+x^2$, so the left side becomes $169-26x$.
- Then $169-26x=119$, giving $26x=50$ and $x=\tfrac{25}{13}$.
💡 Subtracting the two equations wipes out the height, so the only unknown left is the horizontal distance.
8.G.B.7 Step 4 Solve for the height
- Put $x=\tfrac{25}{13}$ back into the first equation $x^2+h^2=25$.
- Since $x^2=\tfrac{625}{169}$, we get $h^2=25-\tfrac{625}{169}=\tfrac{4225-625}{169}=\tfrac{3600}{169}$.
- Taking the square root, $h=\tfrac{60}{13}$ (the height is positive).
💡 With the horizontal run known, Pythagoras hands back the vertical height directly.
6.G.A.1 Step 5 Apply the area formula
- The area of a trapezoid is half the sum of the bases times the height.
- The bases add to $52+39=91$, and $91=7\times13$, which cancels the $13$ in the denominator of $h=\tfrac{60}{13}$ cleanly: $\text{Area}=\tfrac{1}{2}\times91\times\tfrac{60}{13}=\tfrac{1}{2}\times7\times60=210$.
- So the area is $210$, choice (C).
💡 Averaging the two base lengths and multiplying by the height gives the trapezoid's area in one stroke.
6.G.A.1 Drop a straight vertical line from $D$ and from $C$ down to the bottom base $AB$ 8.G.B.7 Let the left horizontal foot $AP=x$; then the right foot $QB=13-x$. Both triangl 8.EE.C.7 Both equations contain $h^2$, so subtract the first from the second to make $h^2 8.G.B.7 Put $x=\tfrac{25}{13}$ back into the first equation $x^2+h^2=25$. Since $x^2=\tf 6.G.A.1 The area of a trapezoid is half the sum of the bases times the height. The bases Review
Reasonableness: The height $h=\tfrac{60}{13}\approx4.6$ is shorter than both legs ($5$ and $12$), which must be true because a slanted leg is always at least as long as the vertical gap it spans. The result $210$ sits right in the middle of the answer choices and is a whole number, matching the fact that all five choices are integers. A quick sanity bound: the area is between $\tfrac{1}{2}(91)(4)=182$ and $\tfrac{1}{2}(91)(5)=227.5$ for a height near $4.6$, and $210$ lands squarely inside. Choice (E) $260$ would need height $\tfrac{520}{91}\approx5.7$, larger than the short leg $5$ — impossible — so it is a trap.
Alternative: Extend the legs $DA$ and $CB$ upward until they meet at a point $P$. Because $CD\parallel AB$, triangle $PDC$ is similar to triangle $PAB$ with ratio $CD:AB=39:52=3:4$. From $PD:PA=3:4$ and $PA-PD=DA=5$ we get $PA=20$; from $PC:PB=3:4$ and $PB-PC=CB=12$ we get $PB=48$. Triangle $PAB$ then has sides $20$, $48$, $52$, and since $20^2+48^2=400+2304=2704=52^2$ it is a right triangle with area $\tfrac{1}{2}\cdot20\cdot48=480$. The small triangle $PDC$ has area $(3/4)^2\cdot480=270$, so the trapezoid is $480-270=210$ — the same answer (C).
CCSS standards used (min grade 8)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Splitting the trapezoid into a rectangle plus two right triangles, and finishing with the trapezoid area formula $\tfrac{1}{2}(b_1+b_2)h=210$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Writing $x^2+h^2=25$ and $(13-x)^2+h^2=144$ for the two end triangles, and recovering $h=\tfrac{60}{13}$.)8.EE.C.7Solve linear equations in one variable (Subtracting the two Pythagorean equations to eliminate $h^2$ and solving $169-26x=119$ for $x=\tfrac{25}{13}$.)
⭐ To find a trapezoid's area, drop straight lines from the short base to make right triangles, use the Pythagorean theorem to get the height, then average the two bases and multiply.
⭐ To find a trapezoid's area, drop straight lines from the short base to make right triangles, use the Pythagorean theorem to get the height, then average the two bases and multiply.
More like this
Same archetype — closest grade level first.
