Two concentric circles are of radii 5cm and 3cm

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle Solution: The PQ is a chord of a larger circle and a tangent of a smaller circle. Tangent PQ is Therefore,∠OSP = 90° In ΔOSP ( By the OP 2= OS 2+ SP 2 5 2= 3 2+ SP 2 SP 2= 25 - 9 SP 2= 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 bethe 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: Video Solution: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle 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: • • • •

Transcript Ex 10.2,7 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Given: Let two concentric circles be C1 & C2 with center O AB be chord of the larger circle C2 which touches the smaller circle C1 at point P To find: Length of AB Solution: Connecting OP, OA and OB OP = Radius of smaller circle = 3 cm OA = OB = Radius of larger circle = 5 cm Since AB is tangent to circle C1 OP ⊥ AB ∴ ∠ OPA = ∠ OPB = 90° Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 Hence, AB = AP + PB = 4 + 4 = 8 cm Show More

Let O be the common centre of the two concentric circle. Let PQ be a chord of the larger circle which touches the smaller circle at M. Join OM and OP. Since, the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, ∠OMP = 90° Now, In ΔOMP, we have OP 2= OM 2+ PM 2 [Using Pythagoras theorem] ⇒ (5) 2= (3) 2+ PM 2 ⇒ 25 = 9 + PM 2 ⇒ PM 2= 16 ⇒ PM = 4 cm Since, the perpendicular from the centre of a circle to a chord bisects the chord. Therefore, PM = MQ = 4 cm ∴ PQ = 2 PM = 2 x 4 = 8 cm Hence, the required length = 8 cm. Solution: Let the two concentric circles with centre O. AB be the chord of the larger circle which touches the smaller circle at point P. ∴ AB is tangent to the smaller circle to the point P. ⇒ OP ⊥ AB By Pythagoras theorem in ΔOPA, OA 2= AP 2+ OP 2 ⇒ 5 2=AP 2+ 3 2 ⇒AP 2= 25 - 9 ⇒ AP = 4 In ΔOPB, Since OP ⊥ AB, AP = PB (Perpendicular from the center of the circlebisects the chord) AB = 2AP = 2× 4 = 8 cm ∴ The length of the chord of the larger circle is 8 cm.