| I. \[QB=RC\] |
| II. \[AB=BC\] |
| III.\[\angle BAC={{45}^{o}}\] |
| IV. \[AC=PS\] |
A) Only I and TV are correct
B) I, II and III are correct
C) II, III and IV are correct
D) All are correct
Correct Answer: D
Solution :
I. \[QR=RS\] \[\Rightarrow \]\[QR-BR=RS-SC\] \[\Rightarrow \]\[QB=RC\] II. In \[\Delta RCB\] and\[\Delta AQB\] \[\angle R=\angle Q,\,\,QB=RC,\,\,AQ=RB\] \[\therefore \]\[\Delta RCB\equiv \Delta AQB\] \[AB=BC\] III.\[\because \]\[AB=BC\] So,\[\angle BCA=\angle BAC\] \[\angle BCA+\angle BAC={{90}^{o}}\] \[2\angle BAC={{90}^{o}}\] \[\angle BAC={{45}^{o}}\] IV.\[AC||PS\]and\[AC=PS\]
You need to login to perform this action.
You will be redirected in
3 sec
