A) 7.5 cm
B) 5 cm
C) 10 cm
D) 2.5 cm
Correct Answer: A
Solution :
ST. || QR, \[\therefore \]\[\angle PST=\angle PQR\] \[\angle PTS=\angle PRQ\] By AA- similarity \[\Delta PST\tilde{\ }\Delta PQR\] \[\therefore \]\[\frac{PS}{SQ}=\frac{PT}{TR}=\frac{PR-TR}{TR}\] \[\Rightarrow \]\[\frac{6}{9}=\frac{125-x}{x}\] \[[x=TR]\] \[\Rightarrow \]\[\frac{2}{3}=\frac{125-x}{x}\] \[\Rightarrow \]\[2x=37.5-3x\] \[\Rightarrow \]\[5x=37.5\] \[\Rightarrow \]\[x=\frac{37.5}{5}=7.5\,cm\]You need to login to perform this action.
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