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Area

by crackgmat007 » Thu May 07, 2009 10:15 am
In the below figure, by joining B & D and drawing a line from D to AB such that this line is parallel to AE, I got a rectangle and 2 triangles, one of which is a right triangle and the other an isosceles triangle. Based on this, I computed the area of each of the polygon and arrived at an area of 100 for the figure.

Is my approach correct?
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Re: Area

by dtweah » Thu May 07, 2009 1:04 pm
crackgmat007 wrote:In the below figure, by joining B & D and drawing a line from D to AB such that this line is parallel to AE, I got a rectangle and 2 triangles, one of which is a right triangle and the other an isosceles triangle. Based on this, I computed the area of each of the polygon and arrived at an area of 100 for the figure.

Is my approach correct?
Absolutely. You are always allowed to draw a perpendicular line in any geometric figure, wherever it is possible. Since AE is perpendicular to BA, as long as you make the new line parallel to AE, it will be perpendicular to BA as well. However once the line is drawn, you cannot just impose values or assumptions. For example you can't say I make one half of the line 3 and the other half 5, since 5+3 is 8. Otherwise you are in good shape. Yours is the smartest approach to the problem.
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by crackgmat007 » Thu May 07, 2009 1:27 pm
Great. Using that approach I got 100 as the answer, did you also get the same answer?
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by DeepakR » Thu May 07, 2009 7:06 pm
Yes I too got 100.
A1= Area of the rectangle with L=8 and B=2 so A1=16
A2= Area of triangle with base=6 and ht=8 so A2=24
A3= Area of isosceles triangle with c=10 and a=13 where a is the equal side and c is the third side. So A3=(c/4)(sqrt(4a^2-c^2)=60

Total Area = 16 + 24 + 60 = 100

-Deepak
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by crackgmat007 » Thu May 07, 2009 8:53 pm
gr8..tx
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