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In a circle with centre O, AB and CD are two chords such that AB>CD and AB is perpendicular bisector of CD at E. P is a point on CD and when BP is extended it meets the circle at Q. For any point P, the triangle BPE is similar to triangle
option
a) AEC
b) QDP
c) QAB
d) ABC
Read Solution (Total 12)
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- I think answer is QAB
- 11 years agoHelpfull: Yes(9) No(2)
- Answer is QAB as BPE and QAB are similar(AA)... :)
- 11 years agoHelpfull: Yes(5) No(1)
- Ans is a)AEC:
Becoz.. since triangle BPE is a right angled triangle and according to the figure triangle AEC is also right angled which is similar to BPE. - 11 years agoHelpfull: Yes(4) No(2)
- b shld b crrct...
- 11 years agoHelpfull: Yes(3) No(3)
- pls post full solution
- 11 years agoHelpfull: Yes(2) No(0)
- QDP should be the ans
- 11 years agoHelpfull: Yes(2) No(0)
- QAB IS THE WRITRE ANS
- 11 years agoHelpfull: Yes(1) No(0)
- ans plzz.....
- 11 years agoHelpfull: Yes(1) No(0)
- pls tell me correct ans
- 11 years agoHelpfull: Yes(1) No(0)
- please tell me the correct answer............
- 10 years agoHelpfull: Yes(0) No(0)
- c
because
AQB is 90
and EBQ=ABQ - 10 years agoHelpfull: Yes(0) No(0)
- QAB will be the answer
- 10 years agoHelpfull: Yes(0) No(0)
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