In rhombus ABCD, point P lies on segment AD so that BP ⊥ AD, AP = 3, and PD = 2. What is the area of ABCD? (Note: the figure is not drawn to scale.)

2022 AMC 12B Problem 2

Attempt Problem 2 of the 2022 AMC 12B 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 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

The side length of the rhombus is

AD = AP + PD

In right triangle

APB,

find

BP,

then use area

= AD \cdot BP

The side length is

AD = AP + PD = 5,

AB = 5.

In right triangle

APB,

\begin{aligned} BP &= \sqrt{AB^2 - AP^2} \\ &= \sqrt{25 - 9} = 4. \end{aligned}

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.