Difference between revisions of "1966 AHSME Problems/Problem 4"
(New page: Make half of the square's side x. Now the radius of the smaller circle is x, so it's area is pi*x^2 Now find the diameter of the bigger circle. Since half of the square's side is x, the f...) |
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− | + | == Problem == | |
+ | Circle I is circumscribed about a given square and circle II is inscribed in the given square. If <math>r</math> is the ratio of the area of circle I to that of circle II, then <math>r</math> equals: | ||
− | + | <math>\text{(A) } \sqrt{2} \quad \text{(B) } 2 \quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}</math> | |
− | (B) | + | == Solution == |
+ | |||
+ | Make half of the square's side <math>x</math>. Now the radius of the smaller circle is <math>x</math>, so it's area is <math>\pi x^2</math>. | ||
+ | Now find the diameter of the bigger circle. Since half of the square's side is <math>x</math>, the full side is <math>2x</math>. Using the Pythagorean theorem, you get the diagonal to be <math>2\sqrt{2}x</math>. Half of that is the radius, or <math>x\sqrt{2}</math>. Using the same equation as before, you get the area of the larger circle to be <math>2x^2 \pi</math>. Putting one over the other and dividing, you get two as the answer: or <math>\boxed{(B)}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1966|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:09, 14 January 2018
Problem
Circle I is circumscribed about a given square and circle II is inscribed in the given square. If is the ratio of the area of circle I to that of circle II, then equals:
Solution
Make half of the square's side . Now the radius of the smaller circle is , so it's area is . Now find the diameter of the bigger circle. Since half of the square's side is , the full side is . Using the Pythagorean theorem, you get the diagonal to be . Half of that is the radius, or . Using the same equation as before, you get the area of the larger circle to be . Putting one over the other and dividing, you get two as the answer: or .
See also
1966 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.