2022 AMC 10B Problem 2

Below is the professionally curated solution for Problem 2 of the 2022 AMC 10B, 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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Concepts:rhombus Pythagorean Theorem area

Difficulty rating: 870

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.)

$3\sqrt 5$

$10$

$6\sqrt 5$

$20$

$25$

Solution:

Since we have a rhombus, we know $AB = AD = AP + PD = 5.$ And by the Pythagorean Theorem, $AP^2 + BP^2 = AB^2$ We know that $9+ BP^2 = 25.$ $BP = 4.$ The area of a rhombus is $bh,$ so the area is $5\cdot 4 = 20.$

Thus, the answer is D .

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