Question

$AB$ and $AC$ are two chords of a circle such that $AB=AC=6cm$. If radius of the circle is $5cm$, then $BC$ is:

• A. $4.8cm$
• B. $9.6cm$
• C. $2.4cm$
• D. $8.4cm$

Answer 1

The correct option is B $9.6cm$

According to the question:

$\begin{array}{c}fromtheusingpythogorustheorum:\\ \Rightarrow {\left(\frac{BC}{2}\right)}^2=A{B}^2-(r-x{)}^2\\ =36-(r-x{)}^2----\left(i\right)\\ \Rightarrow {\left(\frac{BC}{2}\right)}^2=r^2-x^2------\left(ii\right)\\ \Rightarrow 36-(r-x{)}^2=r^2-x^2\\ \Rightarrow 36-(r^2+x^2-2rx)=r^2-x^2\\ \Rightarrow 36-r^2-x^2+2rx=r^2-x^2\\ \Rightarrow 2rx=2r^2-36\\ \Rightarrow x=\frac{r^2-18}{r}=\frac{25-18}{5}=\frac{7}{5}\\ Now,substitutexinequ\left(ii\right)\\ \Rightarrow {\left(\frac{BC}{2}\right)}^2=r^2-x^2\\ =5^2-{\left(\frac{7}{5}\right)}^2\\ =25-\frac{49}{25}=\frac{625-49}{25}=\frac{576}{25}\\ \Rightarrow \left(\frac{BC}{2}\right)=\sqrt{\frac{576}{25}}=\frac{24}{5}\\ \Rightarrow \frac{BC}{2}=\frac{24}{5}\\ \Rightarrow BC=\frac{48}{5}=9.6cm\\ sothatcorrectoptionisB.\end{array}$