Finding all lengths

For what it's worth, here's a picture. The picture is drawn to scale with D at the origin. The two red lines are the 1/2 and 2/3 dividers. The coordinates of the left ends of the red lines are [MATH](-13.50,37.09)[/MATH] and [MATH](-4.32,11.87)[/MATH]quad.jpg
 
LCKurtz said:
For what it's worth, here's a picture. The picture is drawn to scale with D at the origin. The two red lines are the 1/2 and 2/3 dividers. The coordinates of the left ends of the red lines are [MATH](-13.50,37.09)[/MATH] and [MATH](-4.32,11.87)[/MATH]View attachment 14381
Click to expand...
The question actually asks for the area to be between a thrid and a half of the whole area what does this mean?
 
amathproblemthatneedsolve said:
The question actually asks for the area to be between a thrid and a half of the whole area what does this mean?
Click to expand...
It's been so long since I read the OP that I forgot and thought you wanted between 1/2 and 2/3 instead of 1/3 and 1/2. Here's a picture using the 1/3 and 1/2. The coordinates for the upper red line's left end are [MATH](-22.68,62.31)[/MATH]. Any line between the two reds will satisfy your area requirement.
lines.jpg
 
So basically your line can be any length thats between 3477 or 2318 So you could just make the length so it's 3478 because that's between 1/3 and 1/2 isn't it?
 
I think I said once before that the question is unclear. It just says (assuming you quoted it accurately), "THEY want the area of the triangular section to be between a third and half of the area of the whole field ." It doesn't say what YOU are to do about it. Are you to tell "them" that AE should be between the two values you found, or just suggest a particular value. I think it's wiser to do the former, but I didn't write the problem.
 
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