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Geometry question related to area calculation. Need Help?

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Geometry question related to area calculation. Need Help?

by pardeep_10sharma » Sun Jun 27, 2010 1:19 am
In the figure, each side of square ABCD has length 1, the length of line
Segment CE is 1, and the length of line segment BE is equal to the length
Of line segment DE. What is the area of the triangular region BCE?[/b]

A) 1/3
B)(2^-2 )/4
C) 1/2
d)(2^-2)/2
e)3/4

I am getting C. By similiar triangle concept. i have used property of ratios of area corresponding to sides square for similiar triangles BEC and BCD.

But answer given by test makers is :- D.[/list]
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by albatross86 » Sun Jun 27, 2010 1:49 am
Since BE = DE , in triangle BED, angle EBD = angle EDB

In triangle BCD, since BC = CD, angle CBD = angle CDB

Therefore the angles EBC and EDC are also equal

This means Triangles EBC and EDC have 2 pairs of common sides and one common angle. Thus they must be CONGRUENT.

In an isosceles triangle, the perpendicular (height) always bisects the base.

That means the perpendicular from E to BD bisects BD. BUT BCD is a right angled isosceles triangle too. So its perpendicular also bisects BD. Thus ECF must be a straightline.

BF = FD = BD/2 = root(2)/ 2

CF = root(BC^2 - BF^2) = root(1 - 2/4) = 1/root(2)

So, area of triangle BED = 1/2*EF*BD = 1/2*(CE+CF)*BD = 1/2*(1+1/root(2) )*root(2 )= [root(2) + 1] / 2

Area of triangle BCD = 1/2*CF*BD = 1/2*1/root(2) *root(2) = 1/2

Now the area of triangle BCE = area of triangle ECD since they are congruent

=> Area of triangle BCE = 1/2 ( Area of triangle BED - Area of triangle BCD)
= 1/2* [ {root(2) + 1} /2 - 1/2 ]
= 1/2 * root(2)/2
= root(2)/4

OA is definitely incorrect.
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by gmatmachoman » Sun Jun 27, 2010 2:06 am
albatross86 wrote:Since BE = DE , in triangle BED, angle EBD = angle EDB

In triangle BCD, since BC = CD, angle CBD = angle CDB

Therefore the angles EBC and EDC are also equal

This means Triangles EBC and EDC have 2 pairs of common sides and one common angle. Thus they must be CONGRUENT.

In an isosceles triangle, the perpendicular (height) always bisects the base.

That means the perpendicular from E to BD bisects BD. BUT BCD is a right angled isosceles triangle too. So its perpendicular also bisects BD. Thus ECF must be a straightline.

BF = FD = BD/2 = root(2)/ 2

CF = root(BC^2 - BF^2) = root(1 - 2/4) = 1/root(2)

So, area of triangle BED = 1/2*EF*BD = 1/2*(CE+CF)*BD = 1/2*(1+1/root(2) )*root(2 )= [root(2) + 1] / 2

Area of triangle BCD = 1/2*CF*BD = 1/2*1/root(2) *root(2) = 1/2

Now the area of triangle BCE = area of triangle ECD since they are congruent

=> Area of triangle BCE = 1/2 ( Area of triangle BED - Area of triangle BCD)
= 1/2* [ {root(2) + 1} /2 - 1/2 ]
= 1/2 * root(2)/2
= root(2)/4

OA is definitely incorrect.

Bhai,

OA is definetly correct!!
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by albatross86 » Sun Jun 27, 2010 2:13 am
Bhai,

OA is definetly correct!!
Please share your solution and method. It would really help me, because I can't see anywhere I've gone wrong.

D seems to suggest 2^-2/2 which basically is the same as saying 1/8.

I think it should be 2^1/2 divided by 4.

Here is another extensive thread with the same answer as mine:

https://www.urch.com/forums/gmat-problem ... bce-2.html
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by gmatmachoman » Sun Jun 27, 2010 2:19 am
Ok actually I had powercut at my place so went for a fag! Now bak to Business!!

Try to determine the angle CBE

It will be LCBE = 22.5

Drop a perpendicular from C to BE. So u form a right angled triangle.

Sin 22.5 = MC/BC

BC= 1(given)

SO how do u find Sin 22.5??

formula : Sin(a/2) =sqrt {(1+cos a)/2}

So a=45 and a/2 = 22.5

Sin 22.5 = sqrt{ ( 1+Cos 45)/2}
= sqrt {1+ 1/V2)/2}

Apply this value to get MC.

In the same way find value of MB.

Then again MB*2 is equal to BE.

SO area of triangle = 1/2 * BE * MC

we have all the values of BE & MC..

Apply them and u will see area = 1/8
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by gmatmachoman » Sun Jun 27, 2010 2:21 am
Guys Actually i almost got furiated with this one..but the EGO never lets us sleep..after the strong fag, i got the momentum..and pumped back..and u can see the result..

I ate few steps..but u can solve using a pen and paper..see attachment of the amateur dwg
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by gmatmachoman » Sun Jun 27, 2010 2:36 am
@albatross

I am looking for ur comments..plz share them!!
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by albatross86 » Sun Jun 27, 2010 2:40 am
Hi gmatmachoman. I see you like trigonometry, I love it too, but somehow I thought we never need to really apply it on the GMAT. It can be useful though!

Actually, for fun I tried out your method, but I got my answer again :)

Sin(22.5) = sqrt( 2 - sqrt(2) ) / 2
Cos(22.5) = 1 / [sqrt(2)*sqrt(2 - sqrt(2) )]

So from this we get MC = Sin(22.5) and MB = Cos (22.5) since BC = 1

Therefore area = 1/2 * BE * MC = 1/2 * 2 * MB*MC = MB*MC = Sin(22.5) * Cos (22.5)

= sqrt(2-sqrt(2)) / 2*sqrt(2) * sqrt(2 - sqrt(2) )
= 1/ 2*sqrt(2)

which is actually the same as sqrt(2)/4

Hope that makes sense, and thanks for your alternate method - it was interesting :)
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by outreach » Sun Jun 27, 2010 5:37 am
The below post by ian says all the answer options are wrong

https://www.beatthegmat.com/area-of-tria ... html#62848
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by akhpad » Sun Jun 27, 2010 6:32 am
I also got 1/2√2 = √2/4

gmatmachoman - there mush be some mistake in your calculation.

Method 1
------------

angle CBE = angle CEB = 22.5
angle BCE = 90+45

Area of ∆ BCE = (1/2) * 1 * 1 * sin(90+45) = (1/2) (1/√2) = 1/2√2 = √2/4

Method 2
-----------
F = Mid point of BD
Area of ∆ BFE - Area of ∆ BFC = (1/2) * (1/√2) * (1 + 1/√2) - (1/2) * (1/√2) * (1/√2) = 1/2√2 - 1/4 - 1/4 = 1/2√2 = √2/4
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