Skip to main content

2020 AMC 12A Problem 18

Problem 18 of 25IntermediateGeometry

Quadrilateral ABCDABCD satisfies ∠ABC=∠ACD=90∘,\angle ABC = \angle ACD = 90^\circ, AC=20,AC = 20, and CD=30.CD = 30. Diagonals ACAC and BDBD intersect at point E,E, and AE=5.AE = 5. What is the area of quadrilateral ABCD?ABCD?

Answer choices

Show solution

Solution

Place A=(0,0)A = (0,0) and C=(20,0).C = (20, 0). Since ∠ACD=90∘,\angle ACD = 90^\circ, D=(20,30),D = (20, 30), and E=(5,0)E = (5, 0) because AE=5.AE = 5. Since ∠ABC=90∘,\angle ABC = 90^\circ, BB lies on the circle of radius 1010 centered at (10,0).(10, 0). Line DEDE is (5+t, 2t);(5 + t,\, 2t); substituting gives t2−2t−15=0,t^2 - 2t - 15 = 0, so t=5t = 5 or t=−3.t = -3. For EE to lie between BB and D,D, take t=−3,t = -3, giving B=(2,−6),B = (2, -6), a distance 66 below line AC.AC. Then [ACD]=12⋅20⋅30=300[ACD] = \tfrac12 \cdot 20 \cdot 30 = 300 and [ABC]=12⋅20⋅6=60,[ABC] = \tfrac12 \cdot 20 \cdot 6 = 60, so the total area is 360.360. Thus, D is the correct answer.

More practice

Concepts: coordinate geometry · circle · area decomposition

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.