question_answer

In\[\Delta ABC,a=13cm,b=12cm\]and\[c=5cm\]Then the distance of A from BC is

A)  \[\frac{\sqrt{3}+1}{3\sqrt{3}}cm\]                          

B)  \[\frac{60}{13}cm\]

C)  \[\frac{65}{12}cm\]       

D)         \[\frac{144}{13}cm\]

E)  \[\frac{65}{13}cm\]

Correct Answer: B

Solution :

Since, \[a=13\text{ }cm,\text{ }b=12\text{ }cm,\text{ }c=5\text{ }cm\] Now, \[{{(13)}^{2}}={{(5)}^{2}}+{{(12)}^{2}}\] \[\Rightarrow \]               \[169=169\] So,\[\Delta ABC\]is a right angled triangle. \[\therefore \] Area       \[=\frac{1}{2}\times 5\times 12=30\,c{{m}^{2}}\] Again, area \[=\frac{1}{2}\times BC\times AD\] \[\Rightarrow \]               \[30=\frac{1}{2}\times 13\times AD\] \[\Rightarrow \]               \[AD=\frac{60}{13}cm\]