In the semicircle with center O, which of the following is the best es

Question

https://gmatclub.com/forum/download/file.php?id=42739 In the semicircle with center O, which of the following is the best estimate for the distance between points P and Q? (A) .33 (B) 1.42 (C) 1.73 (D) 2.65 (E) 2.82 image.JPG

abhishekgup · Answer

Correct answer should be E.
Its a right angle triangle with 2 sides being 2 each. Using pythagoras, hypotenuse PQ would be 2.82

chetan2u · Answer

OP =0Q = radius=√(1^2+(√3)^2)=√4=2
OPQ is isosceles right angled triangle so PQ =2√2=2*1.4=2.8

E

PKN · Answer

Yes answer should be E.

OP=OQ=radius=2

hypotenuse=2*\sqrt{2}=2.82  (Approx.)

hibobotamuss · Answer

chetan2u  hi this might seem like a stupid question but for the life of me i can't understand how you got the radius to be 2?

chetan2u · Answer

Hi..
Point P is on circumference and center is origin..
So the radius is the distance of P from origin..
P is given as (-√3,1) that is x is -√3 and y is 1 so it's a right angled triangle at 0 with sides -√3 and 1 and radius is the hypotenuse
 r^2=(-√3)^2+1^2=3+1=4.....r=2

nisthagupta28 · Answer

I am unable to understand how is this a right angled triangle?

Posted from my mobile device

Bunuel · Answer

It is indicated in the image:

Aannie · Answer

Please can someone explain how we took O as origin , its no where written that O is origin in the question ?

nisthagupta28 · Answer

Hi,

Firstly, thank you for responding. 
However, I'm sorry for not being clear.

The triangle POQ is right angled. I am not confused on that point.

But I want to understand, how can we infer if the PP'O- is right angled triangle.

I'm not aware if we can do that, or if there is any theory behind such inference.

Thank you in advance.

chetan2u · Answer

But, where is the PP’O triangle? There are only three points given P, O and Q.

chetan2u · Answer

Hi
O is the point of intersection of X-axis and Y-axis, so it is the origin.

nisthagupta28 · Answer

We applied Pythagoras theorem to find PO =2. Please correct me, but I thought we assumed a triangle PP'O

Posted from my mobile device

bv8562 · Answer

Bunuel  Can we solve this by assuming point Q to be the mirror image of point P?

napolean92728 · Answer

Even I have the same doubt , I tried solving it got point Q as \((\sqrt{3},1)\), and after calculating the distance got it as \(2\sqrt{3}\) but it's not in the answer options so recognized that I'm following the wrong approach, but now I wanna understand why is it wrong? Bunuel can you pls explain.

Thanks in advance.

bv8562

Bunuel Can we solve this by assuming point Q to be the mirror image of point P?

Thanks in advance.

Krunaal · Answer

The line with point Q will have a slope that is negative reciprocal of the line with Point P, line with point Q =>  y = \sqrt{3}x

Substituting that in circles equation  x^2 + y^2 = 4 , you will get Point Q  (1, \sqrt{3})  and  not   (\sqrt{3},1) ; upon solving you will get the distance  2\sqrt{2}

In the semicircle with center O, which of the following is the best es

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In the semicircle with center O, which of the following is the best es

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards