Difference between revisions of "2022 AMC 10A Problems/Problem 10"
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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) } | + | <math>\textbf{(A) }14 \qquad \textbf{(B) }10</math>\sqrt(2)<math> \qquad \textbf{(C) }16 \qquad \textbf{(D) }</math>12\sqrt(2)<math> \qquad \textbf{(E) }18</math> |
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Revision as of 00:49, 12 November 2022
Problem
Problem 10
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?
\sqrt(2)
12\sqrt(2)
