CPhill

AB = 4  ???

                           A

                                       8

B          2            K          3                C

Since AK is an altitude, triangles AKC and AKB  are  right

AK =  sqrt [ AC^2 - CK^2 ]

AK =  sqrt [ 8^2 - 3^2]  =  sqrt [ 64 - 9 ]  =  sqrt 55

AB = sqrt [ AK^2 + BK^2 ]

AB = sqrt [ 55 + 2^2 ] =  sqrt [59]

The  line y = 5  will intersect the y axis at 5

We can find the point where the line 5x + 2y = 10  crosses the x axis if we let y = 0 

5x =10

x = 2  

The region formed is a triangle  with a base of 2 and a height of  5

Area  =   (1/2) 2 * 5  =   5

Draw altitude CG  (it will also pass through H)

Angle ABH = 30

Angle CBG = 30

Angle CGB = 90

So angle GHB = 60

So angle EHC = 60

And angle HEC =90

So  angle HCA  =  180 - 60 - 90  =  180  - 120  = 30

Let the side of the square = s

Let the distance that each runs = x

The first runner runs x  and the remaining side of the square   = s -x

The second runner runs x

So, since we have a right triangle,   the  distance between them ^2    can be represented as

d^2   =  (  s - x)^2 + x^2

d^2 =  x^2 - 2sx + s^2 + x^2         

d^2 = 2x^2 - 2sx + s^2          take the derivative of this function and set to 0

4x - 2s  = 0

4x =  2s

x  =  (2/4)s

x = (1/2) s

d^2  = (s -(1/2)s)^2 + (1/2s)^2

d^2  = (1/2s)^2  + (1/2s)^2

d^2  = (2/4)s^2

d^2 =  (1/2) s

d =   sqrt ( 1/2   s^2)  =   (1/sqrt 2)  s ≈   .707 s

If s = 1, then Bosco's answer is  good !!!

Excellent reasoning, Bosco !!!!

Here is a similar problem : https://web2.0calc.com/questions/help-pls_105

 E                           F

 M P 

 N     H                G

Let EN  be the height

Triangle ENH is right

Since EP : PH  = 1 : 2

Then EN = 1 / ( 1 + 2) * height  = 1/3  height of trapezoid

So height of triangle PGH =  2/3 height of trapezoid

[ EPF] =  (1/2)EF * (1/3) height →   EF * height / 6  =  6  → EF =  36 / height

[ PGH ]  =  (1/2)GH * (2/3) height   =  EF * height  / 3  = 6 →  GH = 18 / height

[ EFGH ]  =  (1/2) height * [ 36 / height + 18 / height ]  =   (1/2) height [ 54 / height ] =

54 / 2  =   27

Here : https://web2.0calc.com/questions/circles_147

Assuming a right triangle ????

AB  = sqrt [ 8^2  -1^2]  = sqrt (63)

Let x = CD       8 - x =  BD

Angle BAC = 90

Angle DAC =  45

sin DAC = sqrt (2) / 2

sin BCA = sqrt (63) / 8

Because AD is a bisector, we have that

CA/CD  = AB/ BD

1/x =  sqrt (63)/(8-x)

8 - x =  x sqrt (63)

x + xsqrt (63) = 8

x ( 1 + sqrt 63) = 8

x =  8 / ( 1 + sqrt 63) = CD

Law of Sines

CD / sin CAD   =  DA /  sin DCA 

[8/ (1 + sqrt 63)] / (sqrt (2) / 2 ]   = DA / [sqrt (63) / 8]

DA = [ sqrt (63) / 8 ] [ 8 / [ 1 + sqrt (63)' / (sqrt (2) /2) 

DA  = [sqrt (63) / (1 + sqrt 63) / (1/sqrt 2]

DA = sqrt (2) * sqrt (63) / (1 + sqrt 63 ) ≈ 1.26 = length of angle busector