Question

The radii of two concentric circles with centre $O$ are $5cm$ and $13cm$ respectively.The line drawn from the point A of outer circle touches the inner circle at point M and line AE intersects the inner circle at the points C and D.
If $AE=25cm$, then find $AD.(A-C-D-E)$

• A. 16 cm
• B. 10 cm
• C. 12 cm
• D. 8 cm

Answer 1

The correct option is A 16 cm

Draw perpendicular bisector of line $AE$ which divides it into two equal parts i.e. $AF$ and $FE$

$⟹AF=\frac{25}{2}=FE$ ........... $\left(i\right)$

We know that $OD=5cm$ and $OE=13cm$ ....... Radius of two circles

Now, $△OFE={90}^o$

$\therefore$ By Pythagoras theorem,

$O{E}^2=O{F}^2+E{F}^2$

$⟹O{F}^2=169-{\left(\frac{25}{2}\right)}^2=169-\frac{625}{4}$

$=\frac{51}{4}$

$\therefore OF=\frac{\sqrt{51}}{2}$ ........... $\left(ii\right)$

Now, $△OFD={90}^o$

$\therefore$ By Pythagoras theorem,

$O{D}^2=O{F}^2+F{D}^2$

$⟹F{D}^2=O{D}^2-O{F}^2$

$=25-\frac{51}{4}=\frac{100-51}{4}=\frac{79}{4}$

$\therefore FD=\frac{7}{2}=FC$ ...........$\left(iii\right)$ $[OF$ bisects chords $AE$ and $CD]$

Now, $CD=DF+FC=\frac{7}{2}+\frac{7}{2}=7cm$

$\therefore CD=7cm$ ......... $\left(iv\right)$

By using the theorem,

(Two concentric circles having centre O. A line $l$ intersect outer circle at $A$ and $B$ and inner circle at $C$ and $D$. If $AB=x$ and $CD=y$ then $AC=BD=\frac{1}{2}(x-y)$)

We get, $AC=DE=\frac{1}{2}(AE-CD)$ .............. $\left(v\right)$

And $AE=AD+DE$

$⟹AE=AD+\frac{1}{2}(AE-CD)$ ......... From $\left(v\right)$

$⟹AD=AE-\frac{1}{2}(AE-CD)$ ......... From $\left(v\right)$

$=25-\frac{1}{2}(25-7)$

$=25-9=16$

Hence, $AD=16cm$