If ABCD is a rectangle, what is the area of the shaded region?
A) 64
B) 82
C) 96
D) 120
E) 150
A) 64
B) 82
C) 96
D) 120
E) 150
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Method : 1
Triangle BAD is a right angled triangle.So using Pythagorean theorem
BD^2 = AB^2 + AD^2
BD = 25 units (The sides are in the ratio 15:20:K = 3:4:k Eureka! 3:4:5 triplet! )
Area of Triangle BDC = 0.5*Area of rectangle
0.5*CE*BD = 0.5*15*20 (Let E be the point of intersection of the perpendicular from the vertex C onto diagonal BD)
0.5*CE*25 = 0.5*15*20 => CE = 12 units
Triangle CEB is a right angled triangle (Let E be the point of intersection of the perpendicular from the vertex C onto diagonal BD).So using Pythagorean theorem
CB^2 = CE^2 + EB^2
EB = 16 units (The sides are in the ratio 12:Y:20 = 3:Y:5 Eureka 3:4:5 triplet! )
Area of the shaded region = Area of triangle CEB = 0.5*EB*EC = 0.5*16*12 = 96IMO C
Method : 2 An easier way !
Triangles CEB and BAD are similiar (AAA similarity : Angle CEB = Angle BAD = 90, Angle EBC = Angle ADB, Angle BCE = Angle DBA)
Area of triangle CEB/Area of triangle BAD = (CB/BD)^2 = (25/20)^2 = 25/16
0.5*20*15/Area of triangle BAD = 25/16
Area of triangle = 96 square units !
