AMC 8 · 2005 · #19

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

pythagorean-theoremperimeterarea-rectangles identify-subproblems ↑ Prerequisites: pythagorean-theoremperimeter

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Trapezoid ABCD has slant side AB = 30, top base BC = 50, slant side CD = 25, and a vertical height of 24 from B down to the base AD. Find the perimeter of ABCD.

Pick an answer.

(A)

180

(B)

188

(C)

196

(D)

200

(E)

204

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Three of the four sides are given; the missing piece is the bottom base AD. Tool #1 (Draw a Diagram) says: add the right auxiliary lines to the picture and let the shape do the work. Dropping a perpendicular from B to AD and from C to AD cuts the trapezoid into a rectangle in the middle and two right triangles on the sides. The middle rectangle has the same length as the top base BC = 50. Each side triangle has the height 24 as one leg and a known slant (30 or 25) as the hypotenuse, so the Pythagorean theorem finishes each foot. The two famous triples 3-4-5 and 7-24-25 then snap the answer into place — no algebra needed.

1STEP 1

Drop perpendiculars from B and C to AD: the trapezoid splits into two right triangles plus rectangle BCFE, where EF = BC = 50.

AD = AE + EF + FD, BE = CF = 24, EF = BC = 50

2STEP 2

Right triangle ABE has hypotenuse AB = 30 and leg BE = 24, so by Pythagoras AE = 18 (the 3-4-5 triple scaled by 6).

AE = √(30² - 24²) = √(324) = 18

3STEP 3

Right triangle CFD has hypotenuse CD = 25 and leg CF = 24, so by Pythagoras FD = 7 — the 7-24-25 triple.

FD = √(25² - 24²) = √(49) = 7

4STEP 4

Assemble AD = AE + EF + FD = 18 + 50 + 7 = 75, then the perimeter is 30 + 50 + 25 + 75 = 180 → (A).

AD = 18 + 50 + 7 = 75; Perimeter = AB + BC + CD + DA = 30 + 50 + 25 + 75 = 180 → (A)

Answer

180

Quick sanity pass. The bottom base AD = 75 must be longer than the top base BC = 50, since the slant sides lean outward in both directions in the picture, and indeed 75 ＞ 50. The two horizontal overhangs add to 18 + 7 = 25, which equals 75 - 50 — the algebra and the picture agree. Both right triangles are standard Pythagorean triples (3-4-5 scaled by 6, and 7-24-25), so no irrational numbers ever appeared, which fits an AMC 8 problem. Finally, the perimeter 30 + 50 + 25 + 75 = 180 matches the smallest answer choice (A), and the next choice (B) 188 would require AD = 83, which is inconsistent with both Pythagorean computations.

💡Key takeaway

Two perpendicular lines turn the trapezoid into a rectangle plus two right triangles. The 3-4-5 and 7-24-25 triples then hand you the missing base in seconds, and the perimeter is just the sum of the four sides.

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