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

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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>\overline{AD}</math>, <math>AP = 3</math>, and <math>PD = 2</math>. What is the area of <math>ABCD</math>? (Note: The figure is not drawn to scale.)
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>\overline{AD}</math>, <math>AP = 3</math>, and <math>PD = 2</math>. What is the area of <math>ABCD</math>? (Note: The figure is not drawn to scale.)


(Figure redrawn 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>
 
<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==
 
<asy>
<asy>
pair A = (0,0);
pair A = (0,0);
Line 20: Line 37:
</asy>
</asy>


<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>
(Figure redrawn to scale.)
 
==Solution==


<math>AD = AP + PD = 3 + 2 =5</math>
<math>AD = AP + PD = 3 + 2 = 5.</math>


<math>ABCD</math> is a rhombus, so <math>AD = AB = 5</math>
<math>ABCD</math> is a rhombus, so <math>AB = AP = 5</math>.


<math>\bigtriangleup APB</math> is a 3-4-5 right triangle, so <math>BP = 4</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>
The area of the rhombus <math>= bh = (AD)(BP) = 5 * 4 = \boxed{\textbf{(D) }20}</math>.


~richiedelgado
~richiedelgado

Revision as of 17:06, 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.)

[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 $= bh = (AD)(BP) = 5 * 4 = \boxed{\textbf{(D) }20}$.

~richiedelgado

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