document.write( "Question 832764: A circle is inscribed in a square which is in turn inscribed of a larger circle. what is the ratio of the area of the larger circle to that of the smaller circle?\r
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Algebra.Com's Answer #502191 by math-vortex(648)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Hi, there--\r\n" );
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document.write( "THE PROBLEM:\r\n" );
document.write( "A circle is inscribed in a square which is in turn inscribed of a larger circle. what is the ratio of the area of \r\n" );
document.write( "the larger circle to that of the smaller circle?\r\n" );
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document.write( "A SOLUTION:\r\n" );
document.write( "You want the ratio of the area of the larger circle to that of the smaller circle.\r\n" );
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document.write( "Sandwiched between these two circles is a square. (See drawing below.)\r\n" );
document.write( "We say that the smaller circle is inscribed in the square.\r\n" );
document.write( "Likewise, the square is inscribed in the larger circle.\r\n" );
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document.write( "AREA OF SMALL CIRCLE\r\n" );
document.write( "In order to find the area of a circle, we need to know its radius. Let's begin with the smaller circle. Do you\r\n" );
document.write( "see that the radius of the smaller circle (AO) is one-half the side length of the square?\r\n" );
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document.write( "Let s be the side length of the square.\r\n" );
document.write( "Then s/2 is the radius of the smaller circle.\r\n" );
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document.write( "An expression for the area of the smaller circle is \"pi%2Ar%5E2=pi%2A%28s%2F2%29%5E2\" or \"%28s%5E2pi%2F4%29\".\r\n" );
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document.write( "AREA OF LARGE CIRCLE\r\n" );
document.write( "Now consider the larger circle. Do you see that the radius of this circle (OB) is one-half the diameter\r\n" );
document.write( "of the square?\r\n" );
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document.write( "To find the diameter of the square, we use the Pythagorean Equation, \"a%5E2%2Bb%5E2=c%5E2\".\r\n" );
document.write( "The length and width of the square are the legs of a right triangle, the diagonal is the hypotenuse.\r\n" );
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document.write( "\"c%5E2=s%5E2%2Bs%5E2\"\r\n" );
document.write( "\"c%5E2=2s%5E2\"\r\n" );
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document.write( "To find c, we take the square root of both sides of the equation.\r\n" );
document.write( "\"c=sqrt%282s%5E2%29\"\r\n" );
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document.write( "Simplify. (The square root of c^2 is c, and the square root of s^2 is s.)\r\n" );
document.write( "\"c=s%2Asqrt%282%29\"\r\n" );
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document.write( "The length of the diagonal of the square is \"s%2Asqrt%282%29\". The radius of the larger circle is half the length\r\n" );
document.write( "of the diagonal, or \"%28s%2Asqrt%282%29%29%2F2\".\r\n" );
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document.write( "An expression for the area of the larger circle is \"pi%2Ar%5E2=pi%2A%28%28s%2Asqrt%282%29%29%2F2%29%5E2\".\r\n" );
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document.write( "We can simplify this expression. (The (sqrt(2))/2 squared is 2/4=1/2.)\r\n" );
document.write( "\"pi%2A%28%28s%2Asqrt%282%29%29%2F2%29%5E2=%28pi%2As%5E2%29%2F2\"\r\n" );
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document.write( "RATIO OF AREAS\r\n" );
document.write( "Let R be ratio of the area of the larger circle to the area of the smaller circle.\r\n" );
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document.write( "\"R=%28%28s%5E2%2Api%29%2F2%29%2F%28%28s%5E2%2Api%29%2F4%29\"\r\n" );
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document.write( "Simplify. The s^2 terms and pi cancel out.\r\n" );
document.write( "\"R=%28%28cross%28s%5E2%2Api%29%29%2F2%29%2F%28%28cross%28s%5E2%2Api%29%29%2F4%29\"\r\n" );
document.write( "\"R=%281%2F2%29%2F%281%2F4%29\"\r\n" );
document.write( "\"R=2\"\r\n" );
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document.write( "The ratio of the area of the larger circle to the area of the smaller one is 2:1. In other words the larger \r\n" );
document.write( "circle has twice the area of the smaller one. Furthermore, since we use s for the side length of the square, \r\n" );
document.write( "we know that this ratio is true for a square of any side length.\r\n" );
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document.write( "Hope this helps! Feel free to email if you have any questions about the solution.\r\n" );
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document.write( "Good luck with your math,\r\n" );
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document.write( "Mrs. F\r\n" );
document.write( "math.in.the.vortex@gmail.com\r\n" );
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