2022 AMC 10B Problem 2

Problem 2 of 25EasierGeometry

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?

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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 .

Concepts: rhombus · Pythagorean Theorem · area

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