Difference between revisions of "2022 AMC 10B Problems/Problem 2"

Revision as of 15:52, 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.)

(Figure redrawn to scale) $[asy] pair A = (0,0); label("$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$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]$

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

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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Difference between revisions of "2022 AMC 10B Problems/Problem 2"

Revision as of 15:52, 17 November 2022

Problem

Solution

@@ Line 2: / Line 2: @@
 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>
+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>
 <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>