Competition · AMC preparation · step 4 of 4
AMC 8 · 2005 · #19
Grade 8 geometry-2d
Pick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Add the auxiliary lines
Drop perpendiculars from B and C to AD: the trapezoid splits into two right triangles plus rectangle BCFE, where EF = BC = 50.
Drawing perpendicular auxiliary lines is the Grade 4 "identify perpendicular lines in plane figures" move that converts a hard quadrilateral into shapes whose sides we know how to measure.
4.G.A.1Draw A DiagramFind AE on the left
Right triangle ABE has hypotenuse AB = 30 and leg BE = 24, so by Pythagoras AE = 18 (the 3-4-5 triple scaled by 6).
Once the picture shows a right triangle with two known sides, the Grade 8 Pythagorean theorem hands you the third. Recognizing the 3-4-5 family makes the arithmetic instant.
The horizontal stretch of the base lying directly under the slant side AB measures 18.
▸ Why?
That horizontal stretch is one leg of a right triangle whose other leg is the drawn height 24 and whose hypotenuse is the slant side 30, so its length is √(30² - 24²) = √(324).
▸ Why?
In that right triangle the two legs' squares add up to the hypotenuse's square, so the horizontal leg's square is 30² - 24² = 324.
▸ Why?
Once the horizontal leg's square is fixed at 324, the leg itself comes back by undoing the squaring — the positive length whose square is 324 is 18.
Find FD on the right
Right triangle CFD has hypotenuse CD = 25 and leg CF = 24, so by Pythagoras FD = 7 — the 7-24-25 triple.
Same Grade 8 theorem, second triangle. Spotting the 7-24-25 triple is the geometry analogue of recognizing 3-4-5 — pure pattern recall.
8.G.B.7Draw A DiagramAdd up the perimeter
Assemble AD = AE + EF + FD = 18 + 50 + 7 = 75, then the perimeter is 30 + 50 + 25 + 75 = 180 → (A).
Once every side is known, the Grade 3 definition of perimeter — add the side lengths — finishes the problem.
3.MD.D.8Draw A DiagramTwo 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.
- Add the auxiliary lines
- Find AE on the left
- Find FD on the right
- Add up the perimeter
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