In the given figure, the radii of two concentric circles are $12 \space cm$ and $7 \space cm$. $AB$ is the diameter of the bigger circle and $BD$ is a tangent to the smaller circle touching it at $D$. Find the length $AD$. D O B E A

$17.06 \space cm$

Step by Step Explanation:
1. We know that angle in a semicircle is of $90^\circ.$ So, $\angle AEB = 90^\circ.$
2. We also know that in two concentric circles, the chord of the larger circle, which touches the smaller circle, is bisected at the point of contact.
So, $OD \perp BE$ and $OD$ bisects $BE$.
3. Using Pythagoras' theorem in right $\triangle OBD$, we have $$OB^2 = OD^2 + BD^2$$ It is given that $OB = 12 \space cm$ and $OD = 7 \space cm.$ $$\implies BD = \sqrt{OB^2 - OD^2} = \sqrt{(12)^2 - (7)^2} \space cm = \sqrt{ 95 } \space cm$$ Now, $BE = 2BD = 2 \sqrt{ 95 } \space cm. \space \space \space \text{ [ D is the midpoint of BE ] }$
4. Using Pythagoras' theorem in right $\triangle AEB$, we have \begin{aligned} AB^2 = AE^2 + BE^2 && \end{aligned} As, $AB$ is the diameter of the circle, $AB$ = $2 \times OB = 2 \times 12 \space cm = 24 \space cm$ $$\implies AE = \sqrt{AB^2 - BE^2} = \sqrt{(24)^2 - (2 \sqrt{ 95 })^2} \space cm = \sqrt { 196 } \space cm$$
5. Using Pythagoras' theorem in right $\triangle AED$, we have \begin{aligned} AD^2 = AE^2 + DE^2 \end{aligned} We know that $DE = BD = \sqrt { 95 } \space cm \space \space \space \text{[As, OD bisects BE]}$

$\implies AD = \sqrt{AE^2 + DE^2} = \sqrt{ (\sqrt{ 196 })^2 + (\sqrt{ 95 })^2 } \space cm = 17.06 \space cm$