question_answer

In the given figure below, AABC is a right angled triangle with AB = 8 cm, BC = 6 cm. O is the in-centre of the triangle. The radius of the in-circle is

A)  3 cm                           

B)  4 cm

C)  2 cm    

D)  5 cm

Correct Answer: C

Solution :

(c): \[OD=OE=OF=r\] (let) \[\angle CBA={{90}^{{}^\circ }}\] \[\therefore \]\[AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}\] \[=\sqrt{{{8}^{2}}+{{6}^{2}}}=\sqrt{64+~36}\] \[=\sqrt{100}=10\] cm Area of \[\Delta ABC=Area\text{ }of\left( \Delta AOC+\Delta BOC+\Delta AOB \right)\] \[\Rightarrow \]\[\frac{1}{2}\times 6\times 8=\frac{1}{2}AC\times OE+\frac{1}{2}\times BC\times OF+\frac{1}{2}\times AB\times OD\] \[\Rightarrow \]\[24=\frac{1}{2}\times 10\times r+\frac{1}{2}\times 6\times r+\frac{1}{2}\times 8\times r\] \[\Rightarrow \]\[24=5r+3r+4r\] \[\Rightarrow \]\[12r=24\Rightarrow r=\frac{24}{12}=2\]cm.