document.write( "Question 1177065: ABCD is a convex quadrilateral that is inscribed in a circle. AB=10,AD=24, and m∠BAD=90degrees. Find the greatest possible area of ABCD.\r
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Algebra.Com's Answer #805290 by ikleyn(52797) $\"\"$ $\"About$

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document.write( "(1)  From the given data, you can see that the inscribed angle BAD is leaning on a diameter of the circle.\r\n" );
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document.write( "    Thus the diameter length is   =  = 26 units.\r\n" );
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document.write( "(2)  Further, the area of the triangle BAD is   = 5*24 = 120 square units,\r\n" );
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document.write( "     and this area is a fixed value, independent of the position of the vertex C.\r\n" );
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document.write( "(3)  So, only the area of the triangle BCD  depends on the position of the vertex C on the circle.\r\n" );
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document.write( "     Notice that the triangle BCD is ALSO a right angled triangle, having the hypotenuse BC as the diameter.\r\n" );
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document.write( "     The area of this triangle is maximal when its height drawn to the hypotenuse BC is equal to the radius of the circle.\r\n" );
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document.write( "     In other words, the area of this triangle is maximal, when the radius drawn to the vertex C is perpendicular to BC.\r\n" );
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document.write( "(4)  If it is the case, then the area of the triangle BCD is half the product of its base  |BD| = 26 by the radius of   = 13\r\n" );
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document.write( "          =  = 13*13 = 169 square units.\r\n" );
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document.write( "(5)  Thus the maximum area of the quadrilateral ABCD is\r\n" );
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document.write( "          = 120 + 169 = 289  square units.     ANSWER\r\n" );
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