Question

If radii of two concentric circles are $6cm$ and $10cm$, the length (cm) of the chord of the larger circle which is tangent to other is

Answer 1

Let $C_1,C_2$ be two circles of radius $10,6$ respectively.

Let $r_1=10cm$ and $r_2=6cm$

Draw a chord $AB$ tangent to $C_2$ at point $P$.

Join $O-A$ and $O-B$

$OP=6cm$ ....... (Radius of smaller circle)

$OA=OB=10cm$ ....... (Radius of bigger circle)

$AB$ is tangent to $C_2$ and $OP$ perpendicular AB

$\therefore$ $\angle OPA=\angle OPB={90}^o$

Using Pythagoras theorem,

$O{A}^2=O{P}^2+A{P}^2$

$⟹A{P}^2=O{A}^2-O{P}^2$

$⟹A{P}^2={10}^2-6^2=100-36=64$

$\therefore AP=8cm$

Similarly, $PB=8cm$

$\therefore AB=AP+PB=8+8=16cm$

Hence, the answer is $16$.