AMC 10 · 2002 · #25

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation
pythagorean-theoremarea-trianglesinteger-pythagorean-triples identify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 3 insights 📊 Diagram
Problem
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.

Pick an answer.

(A)
182
(B)
195
(C)
210
(D)
234
(E)
260

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

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.

1STEP 1

Drop perpendiculars to make right triangles

Drop verticals from D and C to AB at P and Q: the middle PQCD is a rectangle with PQ=39, so the two end right triangles share 13.

PQ=CD=39, AP+QB=52-39=13
2STEP 2

Name the parts and write Pythagoras

Let AP=x, so QB=13-x. Both triangles rise the same height h, so Pythagoras gives x²+h²=25 and (13-x)²+h²=144.

x²+h²=5²=25 (13-x)²+h²=12²=144
3STEP 3

Subtract to find the horizontal foot

Subtracting the two equations erases h² and x² alike, leaving 169-26x=119, so x=25/13.

(169-26x+x²)-x² = 119 → 169-26x=119 → x=50/26=25/13
4STEP 4

Solve for the height

Back-substitute into x²+h²=25: h²=25-625/169=3600/169, and a height is positive, so h=60/13.

h² = 25-625/169 = 3600/169 → h=60/13
5STEP 5

Apply the area formula

Half the base sum 52+39=91 times 60/13, and 91=7×13 cancels the denominator: 210, choice (C).

Area=1/2(52+39)·60/13=1/2 · 91·60/13=1/2 · 7 · 60=210 → (C)
Answer
210
The height h=60/13≈4.6 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 1/2(91)(4)=182 and 1/2(91)(5)=227.5 for a height near 4.6, and 210 lands squarely inside. Choice (E) 260 would need height 520/91≈5.7, larger than the short leg 5 — impossible — so it is a trap.
💡Key takeaway

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.

  • Drop perpendiculars to make right triangles
  • Name the parts and write Pythagoras
  • Subtract to find the horizontal foot
  • Solve for the height
  • Apply the area formula

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