question_answer

AB is a chord of length 24 cm of a circle of radius 13 cm. The tangents at A and B intersect at a point C. Find the length AC.

A) \[31.2\,cm\]                 

B)        \[12\,cm\]                    

C)        \[28.8\text{ }cm\]       

D)        \[25\text{ }cm\]          

Correct Answer: A

Solution :

Given,                    Chord \[AB=24\text{ }cm,~\] Radius \[OB=OA=13\,cm.\] Draw \[OP\bot AB\] In  \[\Delta \,OPB.\,\,OP\bot AB\] \[\Rightarrow \]            \[AP=PB\] [Perpendicular from centre on chord bisect the chord] \[=\frac{1}{2}AB=12\] Also. \[O{{B}^{2}}=O{{P}^{2}}+P{{B}^{2}}\] \[\Rightarrow \]\[{{(13)}^{2}}=O{{P}^{2}}+P{{B}^{2}}\Rightarrow 169=O{{P}^{2}}+{{(12)}^{2}}\] \[\Rightarrow \] \[O{{P}^{2}}=169-144=25\,\,\Rightarrow OP=5\,cm\] In  \[\Delta \text{ }BPC,\text{ }B{{C}^{2}}={{x}^{2}}+B{{P}^{2}}\][By Pythagoras theorem] \[B{{C}^{2}}={{x}^{2}}+144\]          In   \[\Delta \text{ }OBC,\text{ }O{{C}^{2}}=O{{B}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]            \[{{(x+5)}^{2}}={{(13)}^{2}}+B{{C}^{2}}\] \[\Rightarrow \]            \[x=\frac{288}{10}=28.8\,cm\] Put value of x in (i) \[B{{C}^{2}}={{x}^{2}}+144=\frac{{{(144)}^{2}}}{25}+144=BC=31.2\]\[AC=BC=31.2\text{ }cm\]