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2022 AMC 10B Problems/Problem 2: Difference between revisions

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Created page with "==Problem== In rhombus <math>ABCD</math>, point <math>P</math> lies on segment <math>\overline{AD}</math> so that <math>\overline{BP}</math> <math>\perp</math> <math>\overlin..."
 
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<math>\textbf{(A) }3\sqrt{5}\qquad\textbf{(B) }10\qquad\textbf{(C) }6\sqrt{5}\qquad\textbf{(D) }20\qquad\textbf{(E) }25</math>
<math>\textbf{(A) }3\sqrt{5}\qquad\textbf{(B) }10\qquad\textbf{(C) }6\sqrt{5}\qquad\textbf{(D) }20\qquad\textbf{(E) }25</math>
==Solution==
<math>AD = AP + PD = 3 + 2 =5</math>
<math>ABCD</math> is a rhombus, so <math>AD = AB = 5</math>
<math>\bigtriangleup APB</math> is a 3-4-5 right triangle, so <math>BP = 4</math>.
Area of a rhombus <math>= bh = (AD)(BP) = 5 * 4 = \boxed{\textbf{(D) }20}</math>.


-richiedelgado
-richiedelgado

Revision as of 15:38, 17 November 2022

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.)

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

Solution

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

$ABCD$ is a rhombus, so $AD = AB = 5$

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

Area of a rhombus $= bh = (AD)(BP) = 5 * 4 = \boxed{\textbf{(D) }20}$.


-richiedelgado

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