document.write( "Question 1189795: A circle with radius 3 cm is tangent to sides PQ, PS, and SR of rectangle PQRS, and passes through the midpoint of diagonal PR. The area of the rectangle PQRS, in cm, is \n" ); document.write( "
Algebra.Com's Answer #821272 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Drawing
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\n" ); document.write( "Additional points are
  • A = midpoint of diagonal PR, and on the circle
  • B = midpoint of side PS, and on the circle
  • C = center of the circle
  • D = point directly above point C, on the circle and side PQ
  • E = point directly below point C, on the circle and side SR
Points B, D and E are points of tangency.\r
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\n" ); document.write( "\n" ); document.write( "The somewhat inscribed circle has radius 3 cm.
\n" ); document.write( "The diameter is 2*3 = 6 cm, which is the distance from D to E. Therefore, PS = 6 cm as well.\r
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\n" ); document.write( "\n" ); document.write( "Furthermore, AB = 6 because all diameters of a circle are the same length.
\n" ); document.write( "Point A is halfway between P and Q in terms of horizontal distance. Which indicates that PQ = 2*6 = 12 cm.\r
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\n" ); document.write( "\n" ); document.write( "We found that
\n" ); document.write( "PS = 6
\n" ); document.write( "PQ = 12
\n" ); document.write( "Therefore the area of the rectangle is
\n" ); document.write( "area = length*width
\n" ); document.write( "area = PS*PQ
\n" ); document.write( "area = 6*12
\n" ); document.write( "area = 72\r
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\n" ); document.write( "\n" ); document.write( "Answer: 72 square cm
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