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Area of a regular octagon is 4(root 2 + 1) square units. Find its side.

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@Taufique bhai, please help.
Let the side of octagon be "x" units.

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Other answers:

\[4\sqrt{2+1}\] or \[4\sqrt{2}+1\]
|dw:1334303032758:dw| @fractal5 , the first one
are you sure, because that is = \[4\sqrt{3}\]
Now, |dw:1334303202974:dw|
The actual area is :|dw:1334303237350:dw|
Area = ( 2 + 2√2 )*(x^2) where x is the side length
in the regular octagon.there will be 8 equilateral triangle hence let the side is x. then area of octagon=8*area of one equilateral triangle =\[8\times \sqrt{3}\div 4 (x ^{2})\]..find x.
@Taufique bhai, can u say the answer?
Are you sure they are equilateral? I think in a hexagon they are equil. not octagon
Even I am in doubt whther we will get 8 equilateral traingles or not.
Just solve the formula I gave
@fractal5 , u r rite. I got it. Can u say how u got it?
@fractal5,can u plz explain the formula how u got it?
@gurvinder bhai, can u explain the formula?
@Aadarsh fractal is right .. |dw:1334303786166:dw|
Okay. But can u explain the formula which @fractal5 used? plzzzzzzzz
@Taufique bhai, u der?
|dw:1334304124446:dw| find its area and multiply it by 8 and put it equal to given area in your question...
But, now can u explain the nest step? The height and side in terms of "x"?
Aadarsh wait.
Okay bhai. I will be coming in 15 minutes after lunch.
|dw:1334304950665:dw| put the value of tan67.5 deg equal to
Okay bhai. But how to obtain value of tan 67.5?
tan67.5deg=root2+1 put there and find x
@Aadarsh ab kho bhai koi problem.. sorry kam kar rha tha college ka project tha.
Aadarsh khan ho.
|dw:1334305876442:dw| find tan67.5 from here.
But abhi isko solve kaise karenge?
tan67.5=x let |dw:1334306229613:dw| x=root2+1 is only possible because 67.5 is in first quadrant and in first quadrant tan ins positive.hence neglect negative value(1-root2)
OH yeah I got it. Thanks bhai!!!!!!!
ji han vai apko v dhanyabad samajne k liye..
Thanks ji.

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