In the given figure, AB and CD are the parallel chords of a circle with centre O. Such that AB = 8 cm and CD = 6 cm. If OM ⊥ AB and OL ⊥ CD distance between LM is 7cm. Find the radius of the circle?  Let OM be x. ∴ OL = 7 – x In the right ΔAOM, OA2 = AM2 + OM2 = 42 + x2 OA2 = 16 + x2 r2 = 16 + x2 ...(1) [r is the radius] In the right ΔOCL, OC2 = OL2 + CL2 r2 = (7 – x)2 + 32 = 49 + x2 – 14x + 9 = 58 + x2 – 14x …….. (2) From (1) and (2) we get, 16 + x2 = 58 + x2 – 14x 14x = 58 – 16 14x = 42 x = `42/14` x = 3 cm r2 = 16 + x2 = 16 + 9 = 25 ∴ r = `sqrt(25)` = 5 ∴ radius of the circle = 5 cm. Concept: Theorem: If One Side of a Cyclic Quadrilateral is Produced Then the Exterior Angle is Equal to the Interior Opposite Angle. Is there an error in this question or solution? AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD= 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is
\[3\sqrt{5}cm\] Let distance between the centre and the chord CD be x cm and the radius of the circle is r cm.
In triangle OND, `x^2 + 36 = r^2` …… (1) Now, in triangle AOM, `r^2 = 9 + (x +3)^2` …… (2) From (1) and (2), we have, `r^2 = 9 + (sqrtr^2 - 36 + 3)^2` `⇒ r^2 = 9 + r^2 - 36 + 9 + 6 sqrt(r^2 - 36 )` `⇒ 3 = sqrt(r^2 - 36)` `⇒9 = r^2 - 36` `⇒r^2 = 45 ⇒ r =3sqrt(5)` Concept: Angle Subtended by a Chord at a Point Is there an error in this question or solution? Description for Correct answer: Given,
Therefore, PQ = PO+OQ = 3+4 = 7 cm Part of solved Geometry questions and answers : >> Elementary Mathematics >> Geometry Comments Similar Questions
(i) Consider AB and CD as the two chords of the circle where AB || CD lying on the same side of the circle It is given that AB = 8cm and OB = OD = 5cm Draw a line to join the points OL and LM The perpendicular from the centre of a circle to a chord bisects the chord We know that LB = ½ × AB By substituting the values we get LB = ½ × 8 So we get LB = 4cm We know that MD = ½ × CD By substituting the values we get MD = ½ × 6 So we get MD = 3cm Consider △ BLO Using the Pythagoras theorem it can be written as OB^2 = LB^2 + LO^2 By substituting the values we get 5^2 = 4^2 + LO^2 On further calculation LO^2 = 5^2 - 4^2 So we get LO^2 = 25 – 16 By subtraction LO^2 = 9 By taking the square root LO = √9 LO = 3cm Consider △ DMO Using the Pythagoras theorem it can be written as OD^2 = MD^2 + MO^2 By substituting the values we get 5^2 = 3^2 + MO^2 On further calculation MO^2 = 5^2 - 3^2 So we get MO^2 = 25 – 9 By subtraction MO^2 = 16 By taking the square root MO = √16 MO = 4cm So the distance between the chords = MO – LO By substituting the values Distance between the chords = 4 – 3 = 1cm Therefore, the distance between the chords on the same side of the centre is 1cm. (ii) Consider AB and CD as the chords of the circle and AB || CD on the opposite sides of the centre It is given that AB = 8cm and CD = 6cm. Construct OL ⊥ AB and OM ⊥ CD Join the diagonals OA and OC We know that OA = OC = 5cm The perpendicular from the centre of a circle to a chord bisects the chord We know that AL = ½ × AB By substituting the values we get AL = ½ × 8 So we get AL = 4cm We know that CM = ½ × CD By substituting the values we get CM = ½ × 6 So we get CM = 3cm Consider △ OLA Using the Pythagoras theorem it can be written as OA^2 = AL^2 + OL^2 By substituting the values 5^2 = 4^2 + OL^2 On further calculation OL^2 = 5^2 - 4^2 So we get OL^2 = 25 – 16 By subtraction OL^2 = 9 By taking the square root OL = √9 OL = 3cm Consider △ OMC Using the Pythagoras theorem it can be written as OC^2 = OM^2 + CM^2 By substituting the values 5^2 = OM^2 +3^2 On further calculation OM^2 = 5^2 - 3^2 OM^2 = 25 – 9 By subtraction OM^2 = 16 By taking the square root OM = √16 OM = 4cm Distance between the chords = OM + OL So we get Distance between the chords = 4 + 3 = 7cm Therefore, the distance between the chords on opposite sides of the centre is 7 cm. |
In the given figure AB and CD are two parallel chords of a circle with centre O and radius 5cm
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