Question

In a circle of radius $5cm$ and $AB$ and $AC$ are two chords such that $AB=AC=6cm$. The distance of the chord $BC$ from the center will be equal to:

• A. $3.6cm$
• B. $1.4cm$
• C. $1cm$
• D. $4cm$

Answer 1

The correct option is B $1.4cm$

Let $OM=x\Rightarrow AM=5-x$

Here $OB=OC=5=R$

As $△BMO$ is right angled triangle

$\Rightarrow O{B}^2=O{M}^2+B{M}^2$

$\Rightarrow B{M}^2=25-x^2$

Also $△AMB$ is a right angled triangle

$A{B}^2=A{M}^2+B{M}^2$

$\Rightarrow 6^2=(5-x{)}^2+{\left(\sqrt{25-x^2}\right)}^2$

$\Rightarrow 36=25+x^2-10x+25-x^2$

$\Rightarrow 10x=14$

$\therefore x=OM=1.4$cm