SOLUTION: 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. No picture is shown in this problem.

(1)  From the given data, you can see that the inscribed angle BAD is leaning on a diameter of the circle.


    Thus the diameter length is   =  = 26 units.



(2)  Further, the area of the triangle BAD is   = 5*24 = 120 square units,

     and this area is a fixed value, independent of the position of the vertex C.



(3)  So, only the area of the triangle BCD  depends on the position of the vertex C on the circle.


     Notice that the triangle BCD is ALSO a right angled triangle, having the hypotenuse BC as the diameter.


     The area of this triangle is maximal when its height drawn to the hypotenuse BC is equal to the radius of the circle.


     In other words, the area of this triangle is maximal, when the radius drawn to the vertex C is perpendicular to BC.



(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


          =  = 13*13 = 169 square units.



(5)  Thus the maximum area of the quadrilateral ABCD is


          = 120 + 169 = 289  square units.     ANSWER