NCERT Solutions Class 9th Maths Chapter – 9 Circles Exercise 9.2

NCERT Solutions Class 9th Maths Chapter – 9 Circles

NCERT Solutions Class 9th Maths Chapter – 9 Circles

Chapter – 9

Circles

Exercise 9.2

Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord. Solution – We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord. ∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord. ∴∠OLP = ∠OLQ = 90° and PL = LQ Now, in right ∆OLP, we have PL2 + OL2 = 2 ⇒ PL2 + (4 – x)2 = 52 ⇒ PL2 = 52 – (4 – x)2 ⇒ PL2 = 25 -16 – x2 + 8x ⇒ PL2 = 9 – x2 + 8x …(i) Again, in right ∆O’LP, PL2 = PO‘2 – LO‘2 = 32 – x2 = 9 – x2 …(ii) From (i) and (ii), we have 9 – x2 + 8x = 9 – x2 ⇒ 8x = 0 ⇒ x = 0 ⇒ L and O’ coincide. ∴ PQ is a diameter of the smaller circle. ⇒ PL = 3 cm But PL = LQ ∴ LQ = 3 cm ∴ PQ = PL + LQ = 3cm + 3cm = 6cm Thus, the required length of the common chord = 6 cm.

Question 2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. Solution – Given: A circle with centre O and equal chords AB and CD intersect at E. To Prove: AE = DE and CE = BE Construction : Draw OM ⊥ AB and ON ⊥ CD. Join OE. Proof: Since AB = CD [Given] ∴ OM = ON [Equal chords are equidistant from the centre] Now, in ∆OME and ∆ONE, we have ∠OME = ∠ONE [Each equal to 90°] OM = ON [Proved above] OE = OE [Common hypotenuse] ∴ ∆OME ≅ ∆ONE [By RHS congruence criteria] ⇒ ME = NE [C.P.C.T.] Adding AM on both sides, we get ⇒ AM + ME = AM + NE ⇒ AE = DN + NE = DE ∵ AB = CD ⇒ 12AB = 12DC ⇒ AM = DN ⇒ AE = DE …(i) Now, AB – AE = CD – DE ⇒ BE = CE …….(ii) From (i) and (ii), we have AE = DE and CE = BE

Question 3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. Solution – Given: A circle with centre O and equal chords AB and CD are intersecting at E. To Prove : ∠OEA = ∠OED Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OE. Proof: In ∆OME and ∆ONE, OM = ON [Equal chords are equidistant from the centre] OE = OE [Common hypotenuse] ∠OME = ∠ONE [Each equal to 90°] ∴ ∆OME ≅ ∆ONE [By RHS congruence criteria] ⇒ ∠OEM = ∠OEN [C.P.C.T.] ⇒ ∠OEA = ∠OED

Question 4. If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure). Solution – Given : Two circles with the common centre O. A line D intersects the outer circle at A and D and the inner circle at B and C. To Prove : AB = CD. Construction: Draw OM ⊥ l. Proof: For the outer circle, OM ⊥ l [By construction] ∴ AM = MD …(i) [Perpendicular from the centre to the chord bisects the chord] For the inner circle, OM ⊥ l [By construction] ∴ BM = MC …(ii) [Perpendicular from the centre to the chord bisects the chord] Subtracting (ii) from (i), we have AM – BM = MD – MC ⇒ AB = CD

Question 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? Solution – Let the three girls Reshma, Salma and Mandip be positioned at R, S and M respectively on the circle with centre O and radius 5 m such that RS = SM = 6 m [Given] Equal chords of a circle subtend equal angles at the centre. ∴ ∠1 = ∠2 In ∆POR and ∆POM, we have OP = OP [Common] OR = OM [Radii of the same circle] ∠1 = ∠2 [Proved above] ∴ ∆POR ≅ ∆POM [By SAS congruence criteria] ∴ PR = PM and ∠OPR = ∠OPM [C.P.C.T.] ∵∠OPR + ∠OPM = 180° [Linear pair] ∴∠OPR = ∠OPM = 90° ⇒ OP ⊥ RM Now, in ∆RSP and ∆MSP, we have RS = MS [Each 6 cm] SP = SP [Common] PR = PM [Proved above] ∴ ∆RSP ≅ ∆MSP [By SSS congruence criteria] ⇒ ∠RPS = ∠MPS [C.P.C.T.] But ∠RPS + ∠MPS = 180° [Linear pair] ⇒ ∠RPS = ∠MPS = 90° SP passes through O. Let OP = x m ∴ SP = (5 – x)m Now, in right ∆OPR, we have x2 + RP2 = 52 RP2 = 52 – x2 In right ∆SPR, we have (5 – x)2 + RP2 = 62 ⇒ RP2 = 62 – (5 – x)2 ……..(ii) From (i) and (ii), we get ⇒ 52 – x2 = 62 – (5 – x)2 ⇒ 25 – x2 = 36 – [25 – 10x + x2] ⇒ – 10x + 14 = 0 ⇒ 10x = 14 ⇒ x = 14/10 = 1.4 Now, RP2 = 52 – x2 ⇒ RP2 = 25 – (1.4)2 ⇒ RP2 = 25 – 1.96 = 23.04 ∴ RP = √23.04= 4.8 ∴ RM = 2RP = 2 x 4.8 = 9.6 Thus, the distance between Reshma and Mandeep is 9.6 m.

Question 6. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at an equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Solution – Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with Centre O such that AS = SD = DA i. e., ∆ASD is an equilateral triangle. Let the length of each side of the equilateral triangle be 2x. Draw AM ⊥ SD. Since ∆ASD is an equilateral triangle. ∴ AM passes through O. ⇒ SM = 1/2 SD = 1/2 (2x) ⇒ SM = x Now, in right ∆ASM, we have AM2 + SM2 = AS2 [Using Pythagoras theorem] ⇒ AM2= AS2 – SM2= (2x)2 – x2 = 4x2 – x2 = 3x2 ⇒ AM = 3x−−√m Now, OM = AM – OA= (3x−−√ – 20)m Again, in right ∆OSM, we have OS2 = SM2 + OM2 [using Pythagoras theorem] 202 = x2 + (√ 3x – 20)2 ⇒ 400 = x2 + 3x2 – 403x−−√ + 400 ⇒ 4x2 = 40 √3x−− ⇒ x = 10√3 m Now, SD = 2x = 2 x 10√3 m = 20√3 m Thus, the length of the string of each phone = 20√3 m

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