2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 20 Hard

Quadrilateral A B C D satisfies \angle A B C=\angle A C D=90^{\circ}, A C=20, and C D=30. Diagonals \overline{A C} and \overline{B D} intersect at point E, and A E=5. What is the area of quadrilateral A B C D?

  • A.

    330

  • B.

    340

  • C.

    350

  • D.

    360

  • E.

    370

Answer:D

Solution 1:

It's crucial to draw a good diagram for this one. Since A C=20 and C D=30, we get [A C D]=300. Now we need to find [A B C] to get the area of the whole quadrilateral. Drop an altitude from B to A C and call the point of intersection F. Let F E=x. Since A E=5, then A F=5-x. By dropping this altitude, we can also see two similar triangles, B F E and D C E. Since E C is 20-5=15, and D C=30, we get that B F=2 x. Now, if we redraw another diagram just of A B C, we get that (2 x)^{2}=(5-x)(15+x). Now expanding, simplifying, and dividing by the GCF, we get x^{2}+2 x-15=0. This factors to (x+5)(x-3). Since lengths cannot be negative, x=3. Since x=3,[A B C]=60. So [A B C D]=[A C D]+[A B C]=300+60=(\mathbf{D}) 360

Solution 2 (Pro Guessing Strats): We know that the big triangle has area 300 . Use the answer choices which would mean that the area of the little triangle is a multiple of 10 . Thus the product of the legs is a multiple of 20 . Guess that the legs are equal to \sqrt{20 a} and \sqrt{20 b}, and because the hypotenuse is 20 we get a+b=20. Testing small numbers, we get that when a=2 and b=18, a b is indeed a square. The area of the triangle is thus 60 , so the answer is (D) 360