Radii of two concentric circles are 13 and 8. | vedclass

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(D) Let the center of the concentric circles be $P$. The radii are $R = 13$ and $r = 8$.$\overline{AB}$ is a diameter of the larger circle,so $AB = 2

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(D) Let the center of the concentric circles be $P$. The radii are $R = 13$ and $r = 8$.$\overline{AB}$ is a diameter of the larger circle,so $AB = 2 	imes 13 = 26$ and $PB = 13$.$\overline{BD}$ is tangent to the smaller circle at $D$,so $\overline{PD} \perp \overline{BD}$ and $PD = 8$.In right-angled $\Delta PDB$,by the Pythagorean theorem:$BD^2 = PB^2 - PD^2 = 13^2 - 8^2 = 169 - 64 = 105$.Let $\overline{BD}$ intersect the larger circle at $C$. Since $\overline{PD} \perp \overline{BC}$,$D$ is the midpoint of chord $\overline{BC}$,so $CD = BD$. Thus,$CD^2 = 105$.In the larger circle,$\angle ACB$ is an angle in a semicircle,so $\angle ACB = 90^\circ$.In $\Delta ABC$ and $\Delta PBD$,$\angle ACB = \angle PDB = 90^\circ$ and $\angle B$ is common. Thus,$\Delta ABC \sim \Delta PBD$ by $AA$ similarity.Therefore,$\frac{AB}{PB} = \frac{AC}{PD} \implies \frac{26}{13} = \frac{AC}{8} \implies 2 = \frac{AC}{8} \implies AC = 16$.In right-angled $\Delta ACD$,by the Pythagorean theorem:$AD^2 = AC^2 + CD^2 = 16^2 + 105 = 256 + 105 = 361$.$AD = \sqrt{361} = 19$.

Radii of two concentric circles are $13$ and $8$. $\overline{AB}$ is a diameter of the circle with the larger radius. $\overline{BD}$ touches the circle with the smaller radius at $D$. Find $AD$.

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