2022 AMC 10B Problem 2

Attempt Problem 2 of the 2022 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10B solutions, or check the answer key.

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In rhombus $ABCD,$ point $P$ lies on segment $\overline{AD}$ so that $\overline{BP} \perp \overline{AD},$ $AP = 3,$ and $PD = 2.$ What is the area of $ABCD?$

$3\sqrt 5$

$10$

$6\sqrt 5$

$20$

$25$

Answer: D

Concepts:rhombus Pythagorean Theorem area

Difficulty rating: 870

Solution:

Since $ABCD$ is a rhombus, $AB=AD=AP+PD=5.$ Right triangle $ABP$ then gives $\begin{aligned}BP&=\sqrt{AB^2-AP^2}\\&=\sqrt{25-9}=4.\end{aligned}$ Thus the rhombus has base $AD=5$ and height $BP=4,$ so its area is $5\cdot4=20.$

Thus, the answer is D .

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