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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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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
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== Problem ==
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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:
  
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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 2sqrt2*x. Half of that is the radius, or xsqrt2. 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
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<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>
  
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(B)
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== Solution ==
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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>.
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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>.
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== See also ==
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{{AHSME box|year=1966|num-b=3|num-a=5}} 
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

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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