please help!!! HARDEST MATHS QUESTION EVER it will make your brain hurt

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bxx81476
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see post bellow for image
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bxx81476
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DEYA
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Justvisited
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What have you been able to make of it so far? Or are you completely stumped at the outset?
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bxx81476
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completely stumped

(Original post by Justvisited)
What have you been able to make of it so far? Or are you completely stumped at the outset?
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bxx81476
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help
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cheerIeader
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OH HUNNY. I've figured it out.

Occams Razor, the simplest solution is usually the right one.

We put down the pen and walk away because there isn't an answer.
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Justvisited
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So, to get the area in terms of r, we need the radius of the semicircle, r, and the length of the rectangle, which is 2r, and its width which we don't yet know, but we can work it out because we know that the whole perimeter of the shape is 250. So, the first thing to do is to set up an expression for this perimeter in terms of r and the unknown width x, and equate it to 250, then rearrange to make x the subject, giving an expression in terms of r. Once we have that, we can use it to get the area which should come out to the value shown.
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bxx81476
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(Original post by cheerIeader)
OH HUNNY. I've figured it out.

Occams Razor, the simplest solution is usually the right one.

We put down the pen and walk away because there isn't an answer.
just help please
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R T
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Tell me the following:

1. how long the perimeter of the fence is.

2. what the perimeter of a semicircle is in terms of r

3. how much leftover perimeter is used for the fence.

4. how long the "long side" of the rectangle is in terms of r

With all of the above you should be almost done with the first part.
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cheerIeader
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(Original post by bxx81476)
just help please
Oh hunny, homework is set for a reason and I'm not about to help you crack an uncrackable code.
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username2998742
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(Original post by cheerIeader)
OH HUNNY. I've figured it out.

Occams Razor, the simplest solution is usually the right one.

We put down the pen and walk away because there isn't an answer.
Actually that's wrong. Occam's razor states that the solution with the least assumptions is usually the correct one.
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Justvisited
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(Original post by Glaz)
Actually that's wrong. Occam's razor states that the solution with the least assumptions is usually the correct one.
You've actually said the same thing in a more precisely defined way. How else would you define "simplest"?
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bxx81476
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im still confused i got (250-4r)/2
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Justvisited
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(Original post by bxx81476)
im still confused i got (250-4r)/2
For the rectangle width or for the overall area? Both of them are bound to involve pi somewhere
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davidthomasjnr
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(Original post by cheerIeader)
OH HUNNY. I've figured it out.

Occams Razor, the simplest solution is usually the right one.

We put down the pen and walk away because there isn't an answer.
Did you get into Harvard with that logic?
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Sataris
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(Original post by Justvisited)
You've actually said the same thing in a more precisely defined way. How else would you define "simplest"?
Crossing a river is a pretty simple prospect if you assume there's a bridge
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(Original post by bxx81476)
im still confused i got (250-4r)/2
The semicircle has a radius r. So it's area is 1/2 * pi * r^2 .

The base of the semicircle has a length of 2r.

The length of the curve in the semicircle is pi * r

If the fence is 250 long, then 250 = semicircle curve + 3 sides of the rectangle.

We can also say that the difference in length between the curved semicircle line and the straight rectangle edge is going to be pi * r - 2 * r which is (pi-2)*r.

is this getting you anywhere towards finding the area?
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(Original post by monkeybananas)
..
Although it's great you have done the question, usually it is better to let people learn for themselves how to do these questions by giving hints and small steps. In future exams, he won't be able to just copy whole solutions so its better he learns step-by-step how to solve these problems.
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Justvisited
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(Original post by Sataris)
Crossing a river is a pretty simple prospect if you assume there's a bridge
We're talking about hypothesising not execution
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