Australia
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 ^@.
Answer:
^@ 17.06 \space cm ^@
- We know that angle in a semicircle is of ^@90^\circ.^@ So, ^@ \angle AEB = 90^\circ. ^@
- 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 ^@. - 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 ] } ^@
- 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 @^
- 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 ^@
