Since |AB|=|BC|,\(x^2 + 2xy + y^2 +x^2 = 2y^2, so 2x^2 +2xy =2x(x+y)= y^2\)
The ratio of the area of triangle BEC to that of triangle ADB is \(y^2/x(x+y) =2\)

side of triangle BCE = a
side of square = (a+b)
two sides of triangle ACF are b and (a+b)

side of equilatrel triangle = \(a*sqrt(2)\)

sum of area of all triangles = area of square

sqrt(3)/4 * 2a^2 + 1/2 a^2 + 2 * 1/2 * b(a+b) = (a+b)^2

simplify this

b = (sqrt(3) - 1)/2 * a
a+b = a * ( sqrt(3) + 1) / 2

area of BCE = 1/2 a^2
area of AFC = 1/2 * b * (a+b)
= 1/2 * a^2 (3-1) / 4

ratio = 2: 1

answer C

time to solve : 7 minutes

Here is how I approached the problem.

AD (side of square) = 1
BE (side of 45:45:90 triangle) = y
CB (side of equilateral triangle) = z

We're looking for the ratio of the area of BEC to ADB, or \(\frac{BEC}{ADB}\)

If area of Tri. BEC = \(\frac{y^2}{2}\) and Tri.ADB = \(\frac{1(1-y)}{2}\) then we have
\(\frac{y^2}{2} * \frac{2}{1-y} = \frac{y^2}{1-y}\)

if y = \(\frac{1}{2}\), then y^2 = 1/4 and 1-y = 1/2, so we get 1/4 : 1/2 or 1/2....that's not an option.

I believe the size of these triangles can vary, meaning we can increase one to decrease the size of another. So, we can't come up with right answer unless they provide us pre-configured info. No?
If not, I guess I would just press Next on this one

would you please explain how you got the side of the equilateral triangle to be \(a*sqrt(2)\)?

if he makes 1 side, (the base) of BCE = a, it's a 45:45:90 triangle, so it has \(1:1:\sqrt{2}\) sides. Therefore, one side of the triangle makes that hypotnuse and you have \(a\sqrt{2}\)

See the picture I drew above. I have the sides of BCE = y and the hypotnuse = z. He labeled his \(a\) rather than y like I did.

The sides are equal, so it's a 45/45/90 triangle. You know the sides are equal because the inscribed triangle is equilateral.

durgesh, can you label which of these goes to which triangle and how you came up with \(\frac{\sqrt{3}}{4}\)?

\(\frac{\sqrt{3}}{4}\) * 2a^2

1/2 a^2

2 * 1/2 * b(a+b)

How are you guys taking wo sides as contant i am not getting it...........as how the triangle is alligned can vary and so is the ratio of the sides....like allen has taken BE and CE as same is there any rule or was it just an asuumption

but how can you decide that it's a 45:45:90 triangle? what if it's a 30:60:90 triangle? how can you be so sure which one it is? for example, if it were a 30:60:90 triangle, than the hypotenuse would be 2a, then what???

There's only one way the triangle can align if it's an equilateral triangle.

That is what i was trying to say is that the answer can vary from person to person..........Plz can somebody expalin it in detail.........

Because we know it is an equilateral triangle inscribe in a square and both the square and triangle share point A. So the sides coming from point A for the triangle towards the other sides fo the square are the same length.

Look at my post above with the drawing in it. I believe that to be a rather accurate picture of this problem (although not drawn to scale or perfect angles).

We know the inscribed triangle is an equialteral (also an iscoceles). In order for the 2 sides to extend out and hit the square, they must touch in the exact same spot (relative to the side) or one side of the triangle would be longer than the other and it would no longer be equilater (or iscoceles).

Even i was feeling that as if you try to distribute the angles what will be the angles of the triangle..........either what i am decepting is wrong or the problem is wrong..........i am going by the figure given by allen

fine, but this only tells me that ADB and AFC are equal and that DB is the same as CF. I still can't see how we choose the triangle to be either 45:45:90 or 30:60:90. I'm really sorry for troubling you, but would you please explain this point?
thanks

I think i have got it in (triangle ABD) angle BAD((90-60)/2)=15,angle ADB=90,angle ABD=75, using these values in triangle BCE angle cbe will be(180-75-60=45) so from here we got it
i suppose rest can be calculated with this

perfect! that fills the gap in the puzzle! i'll now work through my calculations again to see whether i can get the final answer. thanks

I really appreciate it. ok, what a brutal problem this has been! hehe....i managed to get to the correct answer. How about we all try to figure out whether there could be a faster way to solve this? I honestly don't believe that the GMAC guys actually expect us to use all these endless steps to arrive to our answer. There must be a much faster or more abstract way to solve this....hmm...