Difference between revisions of "2022 AMC 10A Problems/Problem 10"

(Problem)
(Problem)
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card and measures the distance between the two closest vertices of these squares to
 
card and measures the distance between the two closest vertices of these squares to
 
be centimeters, as shown below. What is the area of the original index card?
 
be centimeters, as shown below. What is the area of the original index card?
<math>\textbf{(A) }14 \qquad \textbf{(B) }10</math> <math>\sqrt[2]</math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }12<math> </math>\sqrt[2]<math> \qquad \textbf{(E) }18</math>
+
<math>\textbf{(A) }14 \qquad \textbf{(B) }10 </math>\sqrt[2]<math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }12 </math>\sqrt[2]<math> \qquad \textbf{(E) }18</math>

Revision as of 00:51, 12 November 2022

Problem

Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters. Daniel then cuts out equal squares of side 1 cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be centimeters, as shown below. What is the area of the original index card? $\textbf{(A) }14 \qquad \textbf{(B) }10$\sqrt[2]$\qquad \textbf{(C) }16 \qquad \textbf{(D) }12$\sqrt[2]$\qquad \textbf{(E) }18$