Question

Two concentric circles are of diameters $30cm$ and $18cm$. Find the length of the chord of the larger circle which touches the smaller circle.

Answer 1

$FD=30cm,FO=OD=15cm,EC=18cm,EO=OC=9cm$

$OB=OD$ (radius) $O{O}^{\prime }=OC$ (radius)

In $\Delta O{O}^{\prime }B$, By Pythagoras Theorem,

$O{B}^2=O{O}^{\prime 2}+O^{\prime }{B}^2$

${15}^2=9^2+O^{\prime }{B}^2$

$225-81=O^{\prime }{B}^2$

$O^{\prime }B={12}^2$

$O^{\prime }B=12cm$

Similarly, $A{O}^{\prime }=12cm$

$A{O}^{\prime }=O^{\prime }B$

$AB=A{O}^{\prime }+O^{\prime }B$

$AB=O^{\prime }B+O^{\prime }B$

$AB=12+12$

$\therefore AB=24cm$