A) \[16\,cm\]
B) \[14\,cm\]
C) \[15\,cm\]
D) \[17\,cm\]
Correct Answer: A
Solution :
In \[\Delta DBA\]and\[\Delta DCB,\]we have \[\angle BDA=\angle CDB\] [Each \[{{90}^{o}}\]] and \[\angle DBA=\angle DCB\] [Each\[={{90}^{o}}-\angle A\]] \[\therefore \] \[\Delta \,DBA\tilde{\ }\Delta DCB\] [By AA similarity] \[\Rightarrow \]\[\frac{BD}{CD}=\frac{AD}{BD}\]\[\Rightarrow \]\[CD=\frac{B{{D}^{2}}}{AD}=\frac{8\times 8}{4}=16cm\]
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