Let ABCD and FBED be two 2 x 3 rectangles, as shown. The part where the two rectangles overlap is shaded. Find the area of the shaded region.
+129852 The shaded area forms a rhombus
The long diagonal = √[2^2 + 3^2 ] = √13 units
The center of the rhombus is ( 1, 3/2)
The equation of the line containing the long diagonal is
y = (3/2)x
The line perpendicular to this one has the equation
y = (-2/3) (x - 1) + 3/2
y = (-2/3)x + 2/3 +3/2
y = (-2/3)x + 13/6
So point F is (0, 13/6)
And the distance between F and the center of the rhombus = sqrt [ (0- 1)^2 + ( 13/6 - 3/2)^2 ] =
√ [1 + (2/3)^2 ] = √ [ 9 +4 ] / 3 = √13 / 3
So the length of the other diagonal = (2/3)√13
So......the shaded area = (1/2) product of the diagonal lengths = (1/2) *√13 * (2/3)√13 = (13 / 3 ) units^2
