document.write( "Question 38582: A circle has an inscribed right triangle haveing and altitude of 4 and an area of 28. What are the circumference and area of the circle? \n" ); document.write( "
Algebra.Com's Answer #24098 by AnlytcPhil(1806)\"\" \"About 
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A circle has an inscribed right triangle having an altitude of 4 and an area of\r\n" );
document.write( "28.  What are the circumference and area of the circle?\r\n" );
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document.write( "Note.  A triangle has three altitudes, so the problem is not clear.  It should\r\n" );
document.write( "have stated which altitude is 4.\r\n" );
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document.write( "In the case of a right triangle, each of the two legs are altitudes, and the\r\n" );
document.write( "third altitude is drawn from the vertex of the right angle to the hypotenuse.\r\n" );
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document.write( "Regardless of whether the altitude is one of the legs or the altitude drawn\r\n" );
document.write( "from the vertex to the hypotenuse, we will use this fact:\r\n" );
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document.write( "The hypotenuse of any right triangle inscribed in any circle is a diameter.\r\n" );
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document.write( "Case 1.  The altitude given is a leg.  then h = 4. Since A = 28\r\n" );
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document.write( "The area of a triangle is A = bh/2\r\n" );
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document.write( "                         28 = b(4)/2\r\n" );
document.write( "                         28 = b(2)\r\n" );
document.write( "                         14 = 2b\r\n" );
document.write( "                          7 = b\r\n" );
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document.write( "So the hypotenuse is Ö4² + 7² = Ö16 + 49 = Ö65 \r\n" );
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document.write( "The hypotenuse is a diameter, so the radius is Ö65/2\r\n" );
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document.write( "The circumference of a circle is C = 2pr = 2p(Ö65/2) = pÖ65\r\n" );
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document.write( "The area of a circle is pr² = p(Ö65)² = p(65) = 65p\r\n" );
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document.write( "Case 2.  The altitude given is the one from the vertex of the right angle to\r\n" );
document.write( "the hypotenuse.  Then h = 4.\r\n" );
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document.write( "The area of a triangle is A = bh/2\r\n" );
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document.write( "                         28 = b(4)/2\r\n" );
document.write( "                         28 = b(2)\r\n" );
document.write( "                         14 = 2b\r\n" );
document.write( "                          7 = b\r\n" );
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document.write( "So the hypotenuse is the base b = 7        \r\n" );
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document.write( "The hypotenuse is a diameter, so the radius is 7/2\r\n" );
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document.write( "The circumference of a circle is C = 2pr = 2p(7/2) = 7p\r\n" );
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document.write( "The area of a circle is pr² = p(7/2)² = p(49/4) = 49p/4\r\n" );
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document.write( "Edwin
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