Triagle ABC is right isosceles and PQR is circle with center O. OQ is

I once got bogged down in the solution to a problem similar to this. Maybe it's not worth "solving"...use the answer choices instead.

Hint: 3 of the answers are totally unreasonable, and of the 2 that remain, I think one is much more in the ballpark than the other.

Even to make an educated guess, what is the remotest possibility? Do you believe these kind of questions could be experimental?

Hey,

This problem just seems difficult but it isn't.

From the figure (refer to the one I have attached), we can calculate OB = \(\sqrt{2}\)

Since we know that OQ = 1

x = BQ = OQ + OB = 1 + \(\sqrt{2}\)

Therefore, \(x^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = 3 + 2\sqrt{2}\)

Thus answer is D.

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How can you prove that OR || PB, and similarly OP || BR? Why should OR be exactly perpendicular to BC?

Circle Property : Any line touching the circle is a tangent and makes a 90 degree angle with radius.

Since both AB and BC touch the circle at just one point, that is, points P and R respectively, they can be considered tangents to the circle.

From the circle property mentioned above, any line drawn from O to AB and from O to BC will form an angle of 90 degrees since OR and OP will be the radius of the circle.

Now, since AB and BC are perpendicular to each other, any lines drawn perpendicular to them (OP and OR) will also be perpendicular to each other.

Thus, OPBR forms a square since all angles are 90 degrees and the length of two adjacent is equal (OP = OR = radius of the circle).

Hence we can conclude the following :

PB = BR = OP = OR = radius of the circle = 1 cm.

Hope this helps. If you still have difficulty in understanding any of the above steps, I'll be more than happy to try and explain further.

Cheers.
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Yeah, any question could be experimental. I doubt there are throw-away or stickier/trickier questions deliberately included in the experimental pool. After all, the GMAC tests the experimental questions with the intent of using them as official questions...

Here's my guessing rationale. You know the diameter of the circle is 2, so x is more than 2 and x^2 must be more than 4. That eliminates ACE.

Now the guessing issue is just how much more than 2 is x? With knowledge of the 45-45-90 triangle/diagonal stuff, a scale drawing, and a critical eye, it's not too much of a stretch to just approximate OB as root(2) = 1.4.

Therefore, x is at least around 1 + 1.4 = 2.4, and x^2 is 5.76-ish, or (D).

Also, (B) is just too close to 4 to be as reasonable a guess as (D).

srihari you are outstanding dude!!!

esledge, your guessing technique is impressive as well. Learning to guess properly is in itself an art and can save some questions for sure!!!!

BO^2 = 1 + 1 = 2

BO = root(2)

OQ = 1

BO + OQ = BQ = {root(2) + 1}

BQ^2 = 2 + 1 + 2*root(2)*1 = 3 + 2root(2)

Answer - D
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I also guessed the answer as D.

First of all the x > 2
So answer > 4

Leave with 2 options:
b) 3 + sqrt(2) = 4.4
d) 3 + 2[sqrt(2)] =6.8

I chose D over B cause there is substantial about of x outside diameter of circle which is 4. After squaring x it will sure be more then 4.4

circle_r=1
imagine a square with side r = 1
this square will have diagonal = 1√2
BQ=x=1√2+r=1+√2
x^2=(1+√2)^2=1^2+√2^2+2(1)(√2)=3+2√2

Ans (D)

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