In question no. 3, The position of vertex S and R are not explicitly mentioned with respect to P and Q. I tried interchanging the position and then tried my hands on the question but the value SR came out to be -1. I suppose the diagram was a part of the question.
Nevertheless good question and a wonderful learning platform. I am hoping some GK gyan too in the coming days. Fingers crossed.
Rahul, the position cannot be reversed as if QR is 6, PS is also 6. Now for R to be on the opposite side, QR has to be more than 9 / (rt2) which is greater than 6. So only one case is possible. This is based on an actual CAT question btw, from long ago. 🙂
GK will probably not make an appearance, for several reasons. (a) there is just too much and no way to decide what is “relevant” (b) the CAT, GRE and GMAT, which are the exams we are focusing on most right now, don’t have GK and (c) neither of us (J & T) are particularly fond of GK 😀 Maybe someday we can commission some guest posts on it, but don’t hold your breath….
In the last semi circle problem, triangles PRQ and RTQ are similar, coz angle is 90 and a side common. So deducing from it is: to be similar one angle and one side is enough ?
And the immediate ration x/6 = 6/9; It will be very helpful if there is any cut short logic in taking side’s ratio.
This post might help you understand the similarity logic, Raeez; if you observe carefully, there is a common angle too (and that, rather than the common side, is what needs to be used)!
‘Now for R to be on the opposite side, QR has to be more than 9 / (rt2)’ . I didn’t get this.
is there any formula for the diagonal of the trapezium which has a connection here?
Not a trapezium logic but a “circle+right angle” logic. If we were to take the 45-45-90 triangle with PQ as hypotenuse, we would get 9/rt2 as the other sides. R on the opposite side => QR exceeds this value.
CAT-holics 
Hi,
In question no. 3, The position of vertex S and R are not explicitly mentioned with respect to P and Q. I tried interchanging the position and then tried my hands on the question but the value SR came out to be -1. I suppose the diagram was a part of the question.
Nevertheless good question and a wonderful learning platform. I am hoping some GK gyan too in the coming days. Fingers crossed.
Rahul, the position cannot be reversed as if QR is 6, PS is also 6. Now for R to be on the opposite side, QR has to be more than 9 / (rt2) which is greater than 6. So only one case is possible. This is based on an actual CAT question btw, from long ago. 🙂
GK will probably not make an appearance, for several reasons. (a) there is just too much and no way to decide what is “relevant” (b) the CAT, GRE and GMAT, which are the exams we are focusing on most right now, don’t have GK and (c) neither of us (J & T) are particularly fond of GK 😀 Maybe someday we can commission some guest posts on it, but don’t hold your breath….
regards
J
Hello sir,
In the last semi circle problem, triangles PRQ and RTQ are similar, coz angle is 90 and a side common. So deducing from it is: to be similar one angle and one side is enough ?
And the immediate ration x/6 = 6/9; It will be very helpful if there is any cut short logic in taking side’s ratio.
Thanks
RZ
This post might help you understand the similarity logic, Raeez; if you observe carefully, there is a common angle too (and that, rather than the common side, is what needs to be used)!
https://crackthecat.wordpress.com/2013/09/12/spotting-similarity-and-congruence/
regards
J
‘Now for R to be on the opposite side, QR has to be more than 9 / (rt2)’ . I didn’t get this.
is there any formula for the diagonal of the trapezium which has a connection here?
Not a trapezium logic but a “circle+right angle” logic. If we were to take the 45-45-90 triangle with PQ as hypotenuse, we would get 9/rt2 as the other sides. R on the opposite side => QR exceeds this value.
regards
J
how can you conclude in last question, that it is isosceles trapezium, plz exlplain how did you take QR= SP?
SR is parallel to QP so adjacent angles supplementary. Cyclic quadrilateral so opposite angles supplementary. Ab socho…
regards
J
this quadrilateral is inside a semicircle, how can it be cyclic?
Please think first!
regards
J
cyclic quads are inside a circle but this is inside semicircle, so thats y i said.. plz correct if m wrong..
And the semicircle is part of a circle. So all 4 points are on the perimeter of that circle!
regards
J