question_answer

In the given figure, an in-circle DEF is circumscribed by the right angled triangle in which AF = 6 cm and EC = 15 cm. Then the difference between CD and BD is

A)  1 cm.   

B)  3 cm.

C)  4 cm.                          

D)  5 cm.

Correct Answer: A

Solution :

(a) \[OE=OF\] (In-radius) \[OF=BD\] (tangents from externai point B) and \[EC=DC=15\]cm.. \[BF=x\] cm. \[AB=\left( 6+x \right)\]cm. \[AC=6+15=21\]cm. \[B{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]\[{{(x+15)}^{2}}={{(6+x)}^{2}}+{{(21)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+30x+225=36+{{x}^{2}}+12x+441\] \[\Rightarrow \]\[30x-12x=441+36-225\] \[\Rightarrow \]\[18x=252\] \[\Rightarrow \] \[x=\frac{252}{18}=14\Rightarrow BD=14\]cm. \[\therefore \]Required difference \[=CD-BD\] \[=15-14=1\]cm.