#5**+1 **

Let P be the intersection of AD and BF.

Let Q be the intersection of BC and DE.

Triangle(ABP) is congruent to triangle(CDQ).

The area of rectangle(ABC) = 2·3 = 6.

I plan to find the area of triangle(ABP), which is also the area of triangle(CDQ), and subtract these areas

from the area of the rectangle to get the area of the overlapping region.

Let the length of AP = x.

Let the length of PB = y.

By the Pythagorean Theorem, AB^{2} + AP^{2} = PB^{2} ---> 2^{2} + x^{2} = y^{2}. ---> y^{2} = 4 + x^{2}

Since BP = PD, AP + PD = 3 ---> x + y = 3 ---> x = 3 - y

By substitution: y^{2} = 4 + (3 - y)^{2} ---> y^{2} = 4 + 9 - 6y + y^{2} ---> 6y = 13 ---> y = 13/6

Since x = 3 - y ---> x = 3 - (13/6) = 5/6

Area of triangle(ABP) = ½·2·(5/6) = 5/6

Area of triangle(CDQ) = 5/6

Area of the overlapping region = 6 - (5/6) - (5/6) = 13/3.

geno3141 Jun 29, 2020