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

Lets say DQ = x, QB = 2x
Diagonal = 3x
Side of the square = 3x/√2
Area = (3x/√2)^2 = 9x^2/2 = 4.5
9x^2 = 9; x =1
DQ = 1; QB = 2 ; Diagonal = 3
Side of square = 3/√2

See the attached figure.
Since BD is diagonal, angle BDC is 45 degree.
If you drop a perpendicular from Q to base CD you get a 45-45-90 triangle.
From here it is just application of Pythagoras theorem.
QC = sqrt {(1/√2)^2 + (2/√2)^2
= √10 / 2

Answer A

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

so you can not apply Pythagoras here.

Regards

jimmy

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.

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..

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?

[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.

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?

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.

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