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EXERCISE 10.4, , Q.1. Two circles of radii 5 cm and 8 cm intersect at two points and the distance, , between their centres is 4 cm. Find the length of the common chord., , Sol. In AAOO’,, AO? = 52 = 25, AO?= 32 = 9 A, 002= 42 = 16 iS, AO? + 00? = 9 + 16 = 25 = AO? *, = ZA0’°0 P, = 90° A,, [By converse of pythagoras theorem] B, , Similarly, ZBO’O = 90', , = ZAO'B = 90° + 9 180°, , = _AOB is a straight line. whose mid-point is O., = AB = (3 + 3) cm = 6 cm Ans., , , , , , Q.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., , Sol. Given : AB and CD are two equal chords of a circle which meet at E., , Q3., , To prove : AE = CE and BE = DE, , Construction : Draw OM 1 AB and ON 1 CD and join OE. Proof :, In AOME and AONE, , OM = ON [Equal chords are equidistant], , OE = OE [Common], , ZOME = ZONE [Each equal to 90°] Db, AOME = AONE [RHS axiom|, , > EM = EN «G) — [CPCT), , Now AB = CD [Given] 7, , 1 1, >= 3 = 30D, => AM = CN ..(ii) [Perpendicular from, , centre bisects the chord], Adding (i) and (ii), we get, EM + AM = EN + CN, , > AE = CE (iii), Now, AB=CD wiv), = AB - AE = CD - AE [From (iii)], , = BE=CD-CE Proved., , 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., , |. Given : AB and CD are two equal chords of a circle which meet at E, , within the circle and a line PQ joining the point of intersection to the, centre., To Prove : ZAEQ = DEQ, , Construction : Draw OL 1 AB and OM 1 CD., Proof : In AOLE and AOME, we have, OL = OM [Equal chords are equidistant], , OE = OE [Common], ZOLE = ZOME [Each = 90°], -. AOLE = AOME [RHS congruence], , , , = ZLEO = ZMEO (CPCT]