I need a solution quick thank you very much, the faster the better, thank you! Have a good day

Guest Feb 24, 2019

#1**+3 **

Well...I don't know how "fast" this is

Let B = (0, 0) A = (0, 2) E = (0, 1) F = (1,0) D = (2, 2)

We can find the equation of the line containing AF as y = -2x + 2

And the equation of the line containing BD is y = 1x

So we can find the x coordinate of H as

x = -2x + 2

3x = 2

x = 2/3

And y = 2/3

So...the area of triangle BHF = (1/2) (BF)(2/3) = (1/2) (1) (2/3) = 1/3

And the equation of the line containing ED is (1/2)x + 1

So....the x coordinate of I can be found as

(1/2)x + 1 = -2x+ 2

(5/2)x = 1

x = 2/5

And the y coordinate = (1/2)(2/5) + 1 = 1/5 + 1 = 6/5

Now....the area of triangle ADE = (1/2)(AE)( AD) = (1/2)((1) (2) = 1

And the area of triangle ADI = (1/2)(height) (AD) = (1/2) (2 - 6/5) (2) = (1/2)(4/5)(2) = 4/5

So...the area of triangle AEI = area of ADE - area of ADI = 1 - 4/5 = 1/5

And the area of triangle ABF = (1/2)(BF)(AB) = (1/2)( 1) (2) = 1

So...the area of BEIH = area of ABF - area of AEI - area of BHF =

1 - (1/5) - (1/3) =

1 - 3/15 - 5/15 =

1 - 8/15 =

7 / 15

CPhill Feb 24, 2019