



 
Rahul, I drew a diagram (not to scale) of what you described. $AB = 9, BC = 40, CD = 28$ and $DA = 15$ centimeters. I joined $A$ to $C$ to partition the quadrilateral into 2 triangles. Triangle $ABC$ is a right triangle and its area is easy to calculate. Also since triangle;e $ABC$ is a right triangle you can use Pythagoras Theorem to find the length of the line segment $CA.$ You are left with triangle $CDA$ and you know the lengths of all three of its sides. To find its area look at the second paragraph in my response to Saba. Penny  


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