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well, areaa is L * w
I'm sorry the enclosed area is 80 meters.
Not the whole thing.
area, is area, it doesnt matter.
wait, the enclosed rectangle, or the right triangle?
The triangle is.
ok, then that is completely different. and there is no way to find L. Because the triangle can be area 80, and you can attach a rectangle that is as long as you like, it wont make a difference.
No, no the whole thing including the triangle is 80 meters. Sorry!
ooohhhh... hmm. so its L*w. AND the formula for a triangle
I just sectioned off the triangle for convenience sake. I knew I had to do something with that.
but there are too many unknowns to solve this..
1/2bh+lw=80 |dw:1332740378299:dw| here 6 is hypotenuse. I dnt thnk u cn multiply 6 by x.
unless the triangle is isosolis, so add the two sides with pythegooream theorem
oh... ya.. sorry. right.. its base*height. but we dont know eiither..
ajprincess, you are correct, it should be x times the square root of 36 - x^2
No I have to use inverse sin or something.... I have to find the area of the triangle and subtract it from 80 to find L
which makes x the square root of 18
we will have two variables x nd L
Liliy, I don't know how you got to square root of 18. You were right when you said there are too many unknowns. If the triangle was isosceles, it could be solved.
But there aren't too many unknowns. There is some law I have to use to find it. This is straight out of the book.
what's the thing?
so make hte triangle arccos(x/6)
but why would that help you?
I found the answer in my notebook but my notes are quite confusing. L = 11.83
You have to divide the triangle up into two smaller triangles and use some weird math.
If that is a rectangle (four right angles), and that is an isosceles right triangle, you can bisect the triangle into two right isosceles triangles. Like so: |dw:1332698201254:dw| 6 * L + .5 * 6 * 3 = 80