A) \[2\]
B) \[\frac{8}{3}\]
C) \[\frac{10}{3}\]
D) \[\frac{11}{3}\]
Correct Answer: B
Solution :
\[Q{{C}^{2}}=PQ\times AQ\] |
\[PQ=\frac{Q{{C}^{2}}}{AQ}=\frac{9}{2}\] |
PC is angle bisector of APQ |
\[\therefore \] \[\frac{PQ}{PR}=\frac{QC}{RC}\] |
\[PR=\frac{PQ\times RC}{QC}=\frac{\frac{9}{2}\times 4}{3}=6\] |
\[R{{C}^{2}}=PR\times BR\]\[\Rightarrow \]\[16=6\times BR\]\[\Rightarrow \]\[BR=\frac{16}{6}=\frac{8}{3}\] |
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