Solution: The chord of the larger circle is a tangent to the smaller circle as shown in the figure below. PQ is a chord of a larger circle and a tangent of a smaller circle. Tangent PQ is perpendicular to the radius at the point of contact S. Therefore, ∠OSP = 90° In ΔOSP (Right-angled triangle) By the Pythagoras Theorem, OP2 = OS2 + SP2 52 = 32 + SP2 SP2 = 25 - 9 SP2 = 16 SP = ± 4 SP is the length of the tangent and cannot be negative Hence, SP = 4 cm. QS = SP (Perpendicular from center bisects the chord considering QP to be the larger circle's chord) Therefore, QS = SP = 4cm Length of the chord PQ = QS + SP = 4 + 4 PQ = 8 cm Therefore, the length of the chord of the larger circle is 8 cm. ☛ Check: NCERT Solutions Class 10 Maths Chapter 10 Video Solution: Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 7 Summary: If two concentric circles are of radii 5 cm and 3 cm, then the length of the chord of the larger circle which touches the smaller circle is 8 cm. ☛ Related Questions: Math worksheets and
Open in App Suggest Corrections 2 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. 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. Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles Is there an error in this question or solution? |