A) the area of \[\Delta \,COD\]
B) twice the area of \[\Delta \,COD\]
C) thrice the area of \[\Delta \,COD\]
D) four times the area of \[\Delta \,COD\]
Correct Answer: D
Solution :
Given, \[AB=2CD\] .....(i) Let PM be perpendicular to AB and CD through O. Since \[\Delta OCD\] and \[\Delta OAB\] are similar, therefore \[\therefore \] \[\frac{Area\,\,\Delta \,AOB}{Area\,\,\Delta \,DOC}=\frac{{{(AB)}^{2}}}{{{(DC)}^{2}}}=\frac{4}{1}\] \[\therefore \]Area \[\Delta \,AOB=4\]Area \[\Delta \,DOC\]You need to login to perform this action.
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