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1966 AHSME Problems/Problem 4: Difference between revisions

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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>.
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
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>.
 
(B)


== See also ==
== See also ==

Latest revision as of 21:09, 14 January 2018

Problem

Circle I is circumscribed about a given square and circle II is inscribed in the given square. If $r$ is the ratio of the area of circle I to that of circle II, then $r$ equals:

$\text{(A) } \sqrt{2} \quad \text{(B) } 2 \quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}$

Solution

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 full side is $2x$. Using the Pythagorean theorem, you get the diagonal to be $2\sqrt{2}x$. Half of that is the radius, or $x\sqrt{2}$. Using the same equation as before, you get the area of the larger circle to be $2x^2 \pi$. Putting one over the other and dividing, you get two as the answer: or $\boxed{(B)}$.

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

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