Difference between revisions of "1975 AHSME Problems/Problem 4"
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+ | ==Problem== | ||
+ | |||
If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second? | If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second? | ||
− | <math>\textbf{(A)}\ 2 \qquad | + | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4 </math> |
− | \textbf{(B)}\ \sqrt2 \qquad | ||
− | \textbf{(C)}\ 1/2 \qquad | ||
− | \textbf{(D)}\ 2\sqrt2 \qquad | ||
− | \textbf{(E)}\ 4 </math> | ||
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Denote the side of one square as <math>s</math>. Then the diagonal of the second square is <math>s</math>, so the side of the second square is <math>\dfrac{s\sqrt{2}}{2}</math>. The area of the second square is <math>\dfrac{1}{2}s^2</math>, so the ratio of the areas is <math>\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{\textbf{(A) } 2}</math>. | Denote the side of one square as <math>s</math>. Then the diagonal of the second square is <math>s</math>, so the side of the second square is <math>\dfrac{s\sqrt{2}}{2}</math>. The area of the second square is <math>\dfrac{1}{2}s^2</math>, so the ratio of the areas is <math>\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{\textbf{(A) } 2}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 15:51, 19 January 2021
Problem
If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
Solution
Solution by e_power_pi_times_i
Denote the side of one square as . Then the diagonal of the second square is , so the side of the second square is . The area of the second square is , so the ratio of the areas is .
See Also
1975 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.