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.

Guest Dec 24, 2019

#1**+2 **

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

CPhill Dec 24, 2019