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2022 AMC 10B Problem 2

2022 AMC 10B Problem 2AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 1 of 5 · about 2 min · position P1-10

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.)

A)
$3\sqrt 5$
B)
$10$
C)
$6\sqrt 5$
D)
$20$
E)
$25$

Answer

Key insight

All sides of a rhombus are equal, so AB = AD = 5, and the 3-4-5 triangle ABP gives height BP = 4; area = base times height = 20.

Solution

Since $P$ lies on $\overline{AD}$ , the side $AD = AP + PD = 3 + 2 = 5$ . A rhombus has four equal sides, so $AB = 5$ as well.

Triangle $ABP$ is right-angled at $P$ with hypotenuse $AB = 5$ and leg $AP = 3$ , so
$BP = \sqrt{5^2 - 3^2} = 4.$

A rhombus is a parallelogram, and $BP$ is the height onto base $AD$ . Therefore the area is $AD \cdot BP = 5 \cdot 4 = 20$ .

The answer is $\boxed{\textbf{(D)}\ 20}$ .

Why this works

A perpendicular from a vertex to the opposite side is exactly the height of a parallelogram, and the equal sides of a rhombus turn the given segment lengths into a right triangle with a known hypotenuse. Recognizing the $3$ - $4$ - $5$ triple finishes it in one step.

The trap

Halving the answer to 10 as if the rhombus were a triangle, or computing a diagonal instead of the height.

Common mistakes

Halving the answer to 10 as if the rhombus were a triangle, or computing a diagonal instead of the height.
Using $AP = 3$ or $PD = 2$ as the side length rather than their sum $5$ .

Techniques

Set up the equation/formula and compute; no special trick needed