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 !

Name the point that intersects line AD from the vertex C as E.

As per Pythagoras theorem -(BA)^2 + (AD)^2 = (BD)^2.
There for by substituting values - BD = 25

Now as per the rule of similarity :
AB/CE=AD/BE=BD/BC.
Substitute the values:
15/CE=20/BE=25/20.

Therefore,if 15/CE=25/20; CE=12.
& if 20/BE =25/20; BE=16.

Now, Area of shaded region - triangle BCE = 1/2 * BE*CE = 1/2*16*12 = 96.

Ans - C

shankar.ashwin wrote:If ABCD is a rectangle, what is the area of the shaded region?

A) 64
B) 82
C) 96
D) 120
E) 150

The area of right triangle BCD = (1/2)(15)(20) = 150.

Since CD=15 and BC=20, right triangle BCD is a multiple of a 3:4:5 triangle:
3:4:5 = 15:20:25.
Thus, BD=25.

If we deem BD the base of triangle BCD, then CE is the corresponding height.
Thus:
(1/2)(25)(CE) = 150
CE = 12.

Since CE=12 and BC=20, right triangle BCE also is a multiple of a 3:4:5 triangle:
3:4:5 = 12:16:20.
Thus, BE=16.

The area of triangle BCE = (1/2)(BE)(CE) = (1/2)(16)(12) = 96.

The correct answer is C.

An important take-away: ALWAYS LOOK FOR SPECIAL TRIANGLES.

In the GMAT if you get to a geometry section with triangles always look for the 3:4:5 rule or the 1:sqrt3:2 rule. You will never go wrong with this rule.

Back to the question:

AD=20 and BA=15. Aha we have a perfect 3:4:5 triangle and BD=25.

As a shortcut to figure out lengths CE and BE, notice that the hypotenuse BC is also a multiple of 5. Again we have a 3:4:5 triangle. Multiply both 3 and 4 by 4 to get the missing sides, 12 and 16.

Now do simple math, (base*height)/2=(12*16)/2=96