2022 AMC 10B Problems/Problem 2

Problem

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.)

$[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy]$

$\textbf{(A) }3\sqrt{5}\qquad\textbf{(B) }10\qquad\textbf{(C) }6\sqrt{5}\qquad\textbf{(D) }20\qquad\textbf{(E) }25$

Solution

$[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); label("$B$", B, NW); pair C = (6,3); label("$C$", C, NE); pair D = (3.75,0); label("$D$", D, SE); pair P = (2.25,0); label("$P$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]$

(Figure redrawn to scale.)

$AD = AP + PD = 3 + 2 = 5$ .

$ABCD$ is a rhombus, so $AB = AP = 5$ .

$\bigtriangleup APB$ is a 3-4-5 right triangle, so $BP = 4$ .

The area of the rhombus $= bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}$ .

~richiedelgado

2022 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2022 AMC 12B (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

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2022 AMC 10B Problems/Problem 2

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