question_answer

ABC is a right angled triangle wit\[\angle B=90{}^\circ ,\] \[a=6\text{ }cm\]. If the radius of the circumcircle is 5 cm, then the area of\[\Delta ABC\]is

A)  \[25c{{m}^{2}}\]   

B)                         \[30c{{m}^{2}}\]

C)  \[36c{{m}^{2}}\]             

D)         \[24c{{m}^{2}}\]

E)  \[48c{{m}^{2}}\]

Correct Answer: D

Solution :

Let D be the centre of circumcircle. \[\therefore \]  \[BD=5\text{ }cm\] In \[\Delta ABC,\]                 \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]               \[100=A{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]               \[A{{B}^{2}}=64\]\[\Rightarrow \]\[AB=8\] \[\therefore \]Area of \[\Delta ABC=\frac{1}{2}\times AB\times BC\]                                 \[=\frac{1}{2}\times 8\times 6=24\,c{{m}^{2}}\]