A) 8cm
B) 9cm
C) 10cm
D) 7cm
Correct Answer: B
Solution :
\[AB||QR\] (given) \[\therefore \] \[\angle PAB=\] Corresp. \[\angle PQR\] \[\angle PBA=\] Corresp. \[\angle PRQ\] \[\angle PAB\] is common to \[\Delta PAB\] and \[\angle PQR\] \[\therefore \] \[\Delta PAB\,\,\tilde{\ }\,\Delta PAB\] (AAAsimilarity theorem) \[\therefore \] \[\frac{PB}{PR}=\frac{AB}{QR}\] \[\Rightarrow \]\[\frac{2}{6}=\frac{3}{QR}\] \[\Rightarrow \]\[2QR=18\] \[QR=9cm.\]You need to login to perform this action.
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