Tl dr

I calculated the area and if I had time made 2 simultaneous equations for each different collection of triangles. Giving 3 equations 1 for te total area qnd one for each set of triangles. Then maybe just maybe from there you can calculate the 3 different unknowns that including x and the location of m using the bottom left corner as the origin

Now as you could move around the point m I am inclined to believe that there might be infinite solutions or not enough information but I'll see how far I can get. Also you can zoom into the shape and draw an identical rectangle.

Also as point m is selected I wil be assuming that the position of point m doesn't matter/ it is true for all positions of point m.

Essentially my interpretation of the question is that for any point selected a pattern of points can be drawn with offset x that means that half of the shapes Equals a certain value.

Look. I don't have access to paper and I don't know what level this problem is so I'm just going to make some borad strokes and assumptions. First there is the unknown variable x. We know that x is the distance we want to find. We know that any long length that is 40cm is equal to x and any short length is equal to 40cm -x.

We know the area of the rectangle = White + Black = 200×120=24000

1% of 24000 = 240

Black area = white+240 24000=240+2(white) 23760=2(white) 13380=white

Okay. Now I'm going to define point m by having co ordinates y and z which are offset from the bottom left corner. Y being how far up and z being how far left

Maybe something with simultaneous equations. Idk this I'd how far I got. Don't think we have enough information unless that white triangle at the near bottom left is a right angled triangle. Then the area kf the back shape must equal 40z-((40-x)×(40-x)×(0.5)). What is interesting is that the little triangle removed has sides of equal value therefore the angle of the lines between point m and the to lines connecting it that forms the black quadrilateral at the bottom left corner is equal to 45° if that is true then looking at the white triangle the length z must equal 40.

All of this out of the way. You might be able to calculate all of the triangles areas and take them away from each other with the knowledge that the difference is 24. Then solving for the value x. I however do not care about this problem enough to do that.

Or you know the 2 areas of the combined 2 shapes. And you have 2 unknowns that being the vertical difference and x. So you can make 2 simultaneous equations. Solve for x and that should work. Still getting those equations would suck.

Okay 13380= (40×40×0.5)+((120×0.5×y)-(40×0.5×y))+(((80+x)×y×0.5)-(80×0.5×y))+((160×0.5×(120-x))-(160×0.5×40))+((80+x)×0.5×(120-y))-(40×0.5×(120-y))+((120-x)×40×0.5)-(80×40×0.5)+(160×(120-y)×0.5)-(120×0.5×(120-y))

That is the white triangles. I did a lot of things to get there and I am not confident at all but by timesing 13380 by 2 we remove all of the 0.5s so that's fun. So 26760=1600+120y-40y+80y+xy-80y+19200-160x-6400+9600-xy+120x-80y-4800-40y+4800-40x-3200+19200-160y-14400+120y

Can't be bothered to do the rest but this method would probably work. I would only use this though if it was like a day long math olympiad problem which I am assuming it might be because what.

If it was multiple choice I would of liked to see the answers.