A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{4}{3}\]
D) \[\frac{8}{3}\]
Correct Answer: D
Solution :
Let \[GD=x\] so that \[AG=2x\] \[\therefore \] \[AD=x+2x=3x\] Since AD is a median, therefore \[BD=\frac{1}{2}\times BC=\frac{1}{2}\times 6=3\] From right angled \[\Delta ABC\] \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\] or \[{{5}^{2}}={{3}^{2}}+{{(3x)}^{2}}\] or \[25=9+9{{x}^{2}}\] or \[9{{x}^{2}}=16\] or \[{{x}^{2}}=\frac{16}{9}\] \[\therefore \] \[x=\frac{4}{3}\] Hence, \[AG=2x\] \[=2\times \frac{4}{3}=\frac{8}{3}\]You need to login to perform this action.
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