In the diagram above, figure ABCD is a square with an area of 4.5 in^2

I got the first half of when you mentioned although post this, I used normal Pythogoras

QD^2 +QC^2 = CD^2
1 +QC^2=9/2
QC => \sqrt{7/2}

what am I doing wrong here??

I seem to get lost here a bit.
What I did first was to get the diagonal DB, since we know the area of 4.5 inch.

area of square = side^2
so the side would be sqrt (4.5)
So Diagonal is sqrt( 4.5 )^2 + sqrt( 4.5 )^2 = sqrt ( 9 ) = 3

Since DQ ratio is 1:2, DQ has to be 1
This is the point I got lost on.

After some reading on another answer I thought the perpendicular line would help, but then i got stuck on how to simplify the square roots in a fraction.
If someone could also help me out on some reading material regarding square roots and fractions and the reciprocals of them that would be great.

Because why is (From Askhere)
QC = sqrt {(1/√2)^2 + (2/√2)^2
= √10 / 2
I mean when raising these to the power you would square root of 3 right?

QC is not perpendicular to DB.(see ratio DQ/QB is 1/2)

so you can not apply Pythagoras here.

Regards

jimmy

[quote="kelvind13"]I got the first half of when you mentioned although post this, I used normal Pythogoras

QD^2 +QC^2 = CD^2
1 +QC^2=9/2
QC => \sqrt{7/2}

what am I doing wrong here??


Hi kelvind13,

To apply Pythagoras theorem it has to be a right triangle.
You have assumed Traingle DQC to be a right triangle which is wrong. This is where you went wrong.
Hope you understood.

MANHATTAN GMAT OFFICIAL SOLUTION:

If the area of the square is 4.5 and the side length is s, then \(s^2=\frac{9}{2}\) and \(s =\frac{3\sqrt{2}}{2}\)

Triangle BCD is a 45–45–90 right triangle, so the ratio of its sides is \(s : s : s \sqrt{2}\). Thus, DB (the hypotenuse) is \(s \sqrt{2}=\frac{3\sqrt{2}}{2}*\sqrt{2}=3\). Furthermore, if DB is 3, and the ratio DQ : QB is 1 : 2, it must be true that DQ = 1 and QB = 2.

To be able to use Pythagorean Theorem, we must create a right triangle with QC as a side by drawing a new line. Such a right triangle will be similar to triangle BCD. Here are two solutions, the only difference being how one draws the extra line:



The correct answer is A.

Hey, can someone pls explain what is wrong with the below approach?

QD^2 +QC^2 = CD^2
1 +QC^2=9/2
QC => \sqrt{7/2}

Hello Chetan,

Can you please explain how did you concluded BT as 1.5?

Dear Bunuel
One thing that I didn't understood from the second Manhattam Official Solution is:
why DQ = 2, as the ratio of both triangles is 1:3?
Could you clarify that to me?
Thanks!

It's a typo there. DQ should be 1.

Hi Bunuel, chetan2u,

Can you please help me to clarify my doubt.

From the given info on the question DB = 3, DQ =1 & QB = 2. Let P be mid point of QB so, DQ = QP = PB = 1. As each side are equal, DQ, QP & PB will subtend an equal angle at C so Angle QCP = Angle QCD = Angle PCB = 30. Now, as per property of Triangle for Triangle QCD \(\frac{QD}{Sin30} = \frac{QC}{Sin45} = \frac{CD}{Sin75}\). By using this logic I am getting a different answer for QC.

Can you please guide as to where I am going wrong.

You have gone wrong in taking all three angles as 30..
If you have got angle QCD as 75 so QCB as 180-75 =105...
Now if I take the centre triangle QCP, it should be isosceles as the side triangles QCD and PCB are similar..
So the two angles become 105*2=210>180.. not possible..

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