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?


A. 1/3

B. \(\frac{\sqrt{2}}{4}\)

C. 1/2

D. \(\frac{\sqrt{2}}{2}\)

E. 3/4

I got sqrt(2) / 4

I am wondering about ans. b and d in fact 2^-2 = 1/4 I think it should be sqrt(2) instead ... maybe I am wrong

Here is how I did it:
let's suppose O the intersection of ABCD diagonals. In tirangle BDE, the points A, C and O are alignes since C is the barycenter of the triangle (CE=CB=CD) and BED is isocel.

We can calculate the area of BCE by substracting the area of BOC from BOE.

area of BOE=BO*BE/2 = (1/sqrt(2)) * (1/sqrt(2) + 1) / 2 = (1 + sqrt(2))/4
area of BOC= 0.25 area of the square= 1/4

Thus area of BCE = (1 + sqrt(2))/4 - 1/4 = sqrt(2)/4

Hello Bunuel
Could you please explain the underlying logic behind highlighted portion. I failed to understand why it would meet at point O. Please explain.
Thanks

another way to do this problem

extend BC to the right, let's say this point is X. Connect E to X so that CXE is a right angle.

Now, for triangle BCE base BC = 1 and height EX = 1/(sqrt)2.
How length of EX is derived:
angle ECX = 180 - angle BCE = 180 - 135 = 45 deg.
so, triangle ECX is a 45-45-90 right angle triangle and EC is 1.

Area = 1/2 * 1 * 1/(sqrt)2 = 1/2*(sqrt)2 = sqrt(2)/4.

HTH

hey ..
why can`t we assume this a 3-d figure where extended part "bedc" seems to be coming out ..
and as a result angle bce=dce=90..and then 1/2 bh
area comes as 1/2..

whats wrong..plz xplain.
thanks

Hi,

I received this question towards the end of my mock test very recently. Since I didn't have enough time to literally solve it, I made an educated guess, used POE and moved on. Fortunately it worked.

Here's how I approached it:

Given:

- ABCD are all equal to 1; therefore BC=1
- CE=1
- BE=DE

Area of BCE = 1/2 x BC x height (ht)

Since ht is the perpendicular distance and CE is the hypotenuse, height<1

Therefore area of BCE = 1/2 x 1 x something less than 1 = LT 1/2------- (this narrows down ur options to A and B)

Now A and B are very close: A= 0.33 and B=0.32.

I chose B (Not much logic here: honestly because every other answer choice had an even denominator and root 2 appeared as a split in 2 ans choices).

It would be great if any expert can help me figure out whether there's any other solid way to eliminate A (other than the super difficult approach already mentioned in the post)

My exam is around the corner and any quick help would be highly appreciated.

Thanks in advance!

My advise is not to use any fancy triangle properties or weird geometry.
This is a 2D geometry question and the shape of the object is a KITE.

This question can be solved under 20 seconds, The trick is use obscure properties of shapes to your advantage ; in this case the area of a KITE.
If you see the question properly it is asking us to find the area of half KITE and then subtract the area of a Half square
Mathematically
Area = 1/2*area of KITE-1/2* area of square
Area = 1/2 (area of KITE-area of square)
Area of the square is super easy .. its 1; half of it will be = 1/2

For finding the area of the KITE you will need its two diagonal. Smaller one is \(\sqrt{2}\). Bigger one is \(1+\sqrt{2}\)
FORMULA FOR KITE AREA= \(\frac{1}{2}\)*(diagonal1 * diagonal 2)

Solve it and you will get \(\frac{\sqrt{2}}{4}\)

_________________

This Q is can be solved if we extend the line from E to touch one of the diagonals of the square.

Diagonals of a square intersect at right angles. This implies angle BPD will be 90.

Bunuel

Hi Bunuel, Can you please explain what property of isosceles triangle is this? I mean if the triangle BED is isosceles and EO is angle bisector, why is it that EO is height of this triangle? I am trying to understand why EO meets at point O, making a 90 degree angle, and not at any other point on the diagonal of square?

Thanks for your help!

Hi Brunuel, how do you get the length of CO to be \frac{\sqrt{2}}{2}?

I thought BC is the hypotenuse. Thus, by pythagoras thm, CO = square root of [1^2 - (sq root 2/2)^2]?

Please help!