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

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

@@ Line 4: / Line 4: @@
 <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

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

Revision as of 15:38, 17 November 2022

Problem

Solution