Difference between revisions of "1966 AHSME Problems/Problem 4"
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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: | ||
| − | Now find the diameter of the bigger circle. Since half of the square's side is x, the full side is 2x. Using the Pythagorean theorem, you get the diagonal to be | + | <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> |
| + | |||
| + | == 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 | ||
(B) | (B) | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1966|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 01:25, 15 September 2014
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
(B)
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. 
