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Question
Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then find the length of BP.
Solution
To find: BP
Construction: Join OP.
`Now ,OA _|_AP`and `OB_|_BP` `[therefore\text{Tangent to a circle is prependicular to the radius through the point of contact}]`
⇒ ∠OAP = ∠OBP = 90°
On applying Pythagoras theorem in ΔOAP, we obtain:
(OP)2 = (OA)2 + (AP)2
⇒ (OP)2 = (8)2 + (15)2
⇒ (OP)2 = 64 + 225
⇒ (OP)2 = 289
`rArr OP=sqrt289`
⇒ OP = 17
Thus, the length of OP is 17 cm.
On applying Pythagoras theorem in ΔOBP, we obtain:
(OP)2= (OB)2 + (BP)2
⇒ (17)2 = (5)2 + (BP)2
⇒ 289 = 25 + (BP)2
⇒ (BP)2 = 289 − 25
⇒ (BP)2 = 264
⇒ BP = 16.25 cm (approx.)
Hence, the length of BP is 16.25 cm.
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