2022 AMC 12B Problem 2

Below is the professionally curated solution for Problem 2 of the 2022 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12B solutions, or check the answer key.

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Concepts:rhombus Pythagorean Theorem area

Difficulty rating: 1020

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 $ABCD?$ (Note: the figure is not drawn to scale.)

$3\sqrt5$

$10$

$6\sqrt5$

$20$

$25$

Solution:

The side length is $AD = AP + PD = 5,$ so $AB = 5.$ In right triangle $APB,$ $BP = \sqrt{AB^2 - AP^2} = \sqrt{25 - 9} = 4.$

Taking $AD$ as the base and $BP$ as the height, the area is $AD \cdot BP = 5 \cdot 4 = 20.$

Thus, the correct answer is D.

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