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April 17th, 2014, 11:35 PM  #1 
Newbie Joined: Apr 2014 From: India Posts: 4 Thanks: 0  Why area differs if approach is changed ?
Hi friends, I am sorry if this not the right section to ask this. I was solving a simple calculus math and was about to get the answer, but got interrupted by my studymate (is it a word ?) and then confusion came like : Figure is like this : A quadrilateral ABCD where AB  CD and AB = 78 CM CD = 52 CM AD = 28 CM BC = 30 CM We need to find the area of this quadrilateral. My approach : Drawing a line CE so, it creates a parallelogram with AECD and AE = 52 CE = 28 CD = 52 AD = 28 From, here I went to the Triangle BCE where BC = 30 CE = 28 (from above) EB = 26 (7852) 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 Now, as I have the distance or height of the quadrilateral ABCD as 336/13 I calculated the area using formula 1/2 (AB+CD) * d (or h) Which is 1680 CM^2, and correct by book itself. But my friend interrupted when I calculated the area of The Triangle BCE and said that as we have the area of Triangle, we have to only calculate the area of the parallelogram and adding both we will get the required area, it sound right to me, so I tried again with his approach where now I had a parallelogram like : AECD, where AE = 52 CE = 28 CD = 52 AD = 28 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 which is not correct as I calculated above, and book also confirmed 1680 CM^2. Any idea why both differs so much (I could understand if it has minor degree varying). 
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)

April 18th, 2014, 04:53 AM  #3 
Newbie Joined: Apr 2014 From: India Posts: 4 Thanks: 0  
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 $\displaystyle ab = ah$. 
April 19th, 2014, 08:14 AM  #6 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  
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  
Senior Member Joined: Apr 2014 From: UK Posts: 967 Thanks: 344  Quote:
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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