Geometry question related to area calculation. Need Help?

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.

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!!

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

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

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

@albatross

I am looking for ur comments..plz share them!!

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

The below post by ian says all the answer options are wrong

https://www.beatthegmat.com/area-of-tria ... html#62848

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