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2022 AMC 12B Problem 2

Problem 2 of 25EasierGeometry

In rhombus ABCD,ABCD, point PP lies on segment AD‾\overline{AD} so that BP‾⊥AD‾,\overline{BP} \perp \overline{AD}, AP=3,AP = 3, and PD=2.PD = 2. What is the area of ABCD?ABCD? (Note: the figure is not drawn to scale.)

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Solution

The side length is AD=AP+PD=5,AD = AP + PD = 5, so AB=5.AB = 5. In right triangle APB,APB, BP=AB2−AP2=25−9=4. \begin{aligned} BP &= \sqrt{AB^2 - AP^2} \\ &= \sqrt{25 - 9} = 4. \end{aligned} Taking ADAD as the base and BPBP as the height, the area is AD⋅BP=5⋅4=20.AD \cdot BP = 5 \cdot 4 = 20. Thus, the correct answer is D.

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Concepts: rhombus · Pythagorean Theorem · area

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