A) 10 cm
B) 12 cm
C) 18 cm
D) 20 cm
Correct Answer: D
Solution :
In \[\Delta \,AOQ\] and \[\Delta \,BOP,\] \[\angle 1=\angle 2\] (vertically opposite angles) \[\angle 3=\angle 4\] (each 90°) \[\therefore \] \[\Delta AOQ\sim \Delta BOP\] (AA similarity criterion) \[\therefore \] \[\frac{AO}{BO}=\frac{AQ}{BP}\] (corresponding sides of similar triangles) \[\Rightarrow \] \[\frac{15}{9}=\frac{AQ}{12}\Rightarrow \,\frac{5}{1}=\frac{AQ}{4}\] \[\Rightarrow \] \[AQ=20\,cm\]You need to login to perform this action.
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