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 April 18th, 2014, 01:24 AM #2 Senior Member   Joined: Apr 2014 From: UK Posts: 967 Thanks: 344 You used the wrong H for the parallelogram, you should have used h (336/13) Thanks from Amanalice
April 18th, 2014, 04:53 AM   #3
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Quote:
 Originally Posted by weirddave You used the wrong H for the parallelogram, you should have used h (336/13)
Ok my bad, but its same H or h, just I wrote H in hurry.

 April 19th, 2014, 07:38 AM #4 Newbie   Joined: Apr 2014 From: India Posts: 4 Thanks: 0 *Bump*
 April 19th, 2014, 08:09 AM #5 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Your friend is wrong. The area of the parallelogram is not 52*28, but in general $\displaystyle area = ab\sin (\textrm{angle between sides})$. Only, if the angle is $\displaystyle 90^{\circ}$, the area is $\displaystyle ab = ah$. Thanks from Amanalice
April 19th, 2014, 08:14 AM   #6
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Quote:
 Originally Posted by Amanalice Then, I calculated the area of this Triangle using S and tried to find the h of this triangle which came like : 336/13 CM
You made a mistake here: $h$ should be $\mathrm{\dfrac{198}{13}\ cm}$. $\mathrm{\dfrac{336}{13}\ cm}$ is the length of EB.

 April 19th, 2014, 09:19 AM #7 Newbie   Joined: Apr 2014 From: India Posts: 4 Thanks: 0 Shit, Now I got it too. Dont know how it got slipped from my logic. Very much thank you friends. Feeling happy
April 19th, 2014, 01:36 PM   #8
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Quote:
 Originally Posted by Amanalice using formula H*W = 52*28 = 1456 CM^2 And, we had already Triangles area as 336 CM^2 And theoretically the area of the quadrilateral should be equal to = 1456+336 =1752 CM^2

The H used was 28, whereas it should have been 336/13 (the same h as the triangle), sorry if it wasn't clear that's what I meant in my previous post.

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