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Japan competition problems, 2007

Geometry Difficulty 4.7 AIME Prove it Japan

Let $ABCD$ be a convex quadrilateral with $AB = 3$ , $BC = 4$ , $CD = 5$ , $DA = 6$ and $\angle ABC = 90^\circ$ . Find the area of $ABCD$ .

Solution

Since $AB = 3$ , $BC = 4$ and $\angle ABC = 90^\circ$ , we get $AC = 5$ . Let $M$ be the midpoint of $AD$ . Because $AM = DM$ , $MC = MC$ , $AC = 5 = DC$ and $\angle ABC = \angle AMC = 90^\circ$ , $AB = AM$ , $AC = AC$ , the triangles $ABC$ , $AMC$ and $DMC$ are all congruent.

Therefore, the area of $ABCD$ is $\frac{3 \times 4}{2} \times 3 = 18$ .

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