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# geometry

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Manager
Joined: 16 Apr 2009
Posts: 230
Schools: Ross

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27 Apr 2009, 19:40
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ABCD is a rectangle.The area of isosceles right triangle ABE =7 and EC=3(BE).AB=BE and AE is within the rectangle The area of the ABCD is?
a)21 b)28 c)42 d)56 e)84

This is my first post.Pls explain this problem...
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Manager
Joined: 02 Mar 2009
Posts: 134

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27 Apr 2009, 22:19

I think E is on the line BC. So you correctly calculated that AB = root(14)--That becomes 1 side

The other side is simply 4*root(14) because of the relationship EC = 3BE

Multiplying the two gives 56.
Manager
Joined: 16 Apr 2009
Posts: 230
Schools: Ross

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28 Apr 2009, 06:07
D is the correct answer. Thanks.
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Manager
Joined: 11 Apr 2009
Posts: 159

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28 Apr 2009, 14:20
Just an alternate explanation;

The area of the triangle ABE is 7. In the rectangle there will be a total of 8 such triangles (since BE= 3EC). Hence the area of the rectangle is 7*8=56.

Director
Joined: 23 May 2008
Posts: 799

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29 Apr 2009, 21:06
forgot about the a=s^2/2 for right isoceles triangle rule, but the key here is to realize that two right isoceles make a square, then you can just say that the area of the square = 14 since half of the sqaure = ABE=7. Area of the square equal side^2 so side=sqrt 14. Area of the rectangle ABCD = side * (side +3side)= 4side^2= 4*14=56

good site

http://www.mathopenref.com/triangle454590.html
Re: geometry   [#permalink] 29 Apr 2009, 21:06
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