Idon't understand how this problem works. special quadrilaterals always confused me

Call the point where the dashed line intersects FG point K.
Given:
.. FK = x+6
.. FG = HI = 3x
.. FI = x+8 = GH = 2x -1
.. Area of a parallelogram = HI*IK
.. Pythagorean theorm says FK^2 +IK^2 = FI^2

Find:
.. x
.. Area

Solution:
.. Using the fact that FI = GH, we find
.. x +8 = 2x -1
.. 9 = x . . . . . . . . . add 1-x to both sides of the equation

Now, we know
.. FK = 9 +6 = 15
.. HI = 3*9 = 27
.. FI = 9 +8 = 17
.. FK^2 +IK^2 = FI^2
.. 15^2 +IK^2 = 17^2
.. Ik^2 = 289 -225 = 64
.. IK = √64 = 8

Area = HI*IK = 27*8 = 216

Jehhnd. rhehbf ehnd new e hw ended nehru shh snow rheh 6 cents!

Answer: b 2 = 1.

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.

therefore, b 2= 1.