MATHS – IX CIRCLES

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## MATHS – IX CIRCLES

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**MATHS – IX**CIRCLES**CIRCLES**Angle Subtended By An Arc Of A Circle**If two chords of a circle are equal ,then their**corresponding arcs are congruent and conversely ,if two arcs are congruent , then their corresponding chords are equal.(in fig 1.1). D A C FIG 1.1 B**Also the angle subtended by an arc at the centre is**defined to be angle subtended by the corresponding chord at the centre in the sense that the minor arc subtends the angle and the major arc subtends reflex angle . Therefore , in fig 1.2 ,the angle subtended by the major arc PQ at o is reflex angle POQ .**FIG:1.2**O Q P**In view of the property above and theorem the following**result is true - Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre . Therefore , the angle subtended by a chord of a circle at its centre is equal to the angles subtended by the corresponding (minor) arc at the centre .**THEOREM**The following theorem gives the relationship between the angles subtended by an arc at a centre and at a point on the circle . THEOREM - The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .**PROOF**Given an arc PQ of a circle subtending angles POQ at the O and PAQ at a point a on the remaining part of the circle . We need to prove that angle POQ = 2 angle PAQ .**CASE- (I)**A O B Q P FIG:1.3**Case -(II)**A P o B Q FIG:1.3**Case – (III)**Q P O B FIG:1.3**Consider the three different cases as given in fig**1.3.In (I) , arc PQ is minor , in (II) , arc PQ is a semicircle and in (III) , arc PQ is major. Let us begin by joining AO and extending it to a point B.**In all the cases ,**^BOQ = ^OAQ + ^AQO Because an exterior angle of a triangle is equal to the sum of the two interior opposite angles.**Also in triangle OAQ ,**OA=OQ (Radii of a circle) Therefore , ^OAQ=^OQA (Theorem) This gives ^BOQ=2^OAQ (I) Similarly , ^BOP=2^OAP (II)**From (I) and (II) , ^BOP +^BOQ = 2(^OAP+**^OAQ) This is same as ^POQ = 2 ^PAQ (III) For the case (III) , where PQ is the major arc , (III) is replaced by reflex angle POQ = 2 ^PAQ.**1.**C If ^AOB =110 degree Then ^ACB =1/2x110 =55 degree. o B A**2 .**A C O If^AOB= 40 degree Then ^ACB=1/2x40 =20 degree. B**3.**If ^AOB =60 degree Then, The reflex angle of AOB= =360-60 =300 degree. B A O**WHAT WE HAVE LEARNT ?**If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment , the four points lie on a circle . OR The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.**EXERCISE**Question 1. Find the value of^AXB.(AOB=40degree) O X (ii) (i) X O B A B A**Answer**1.i If^AOB = 40 degree. Then ^AXB =1/2 x 40 = 20 . Therefore ^AXB =20 degree.**If ^AOB = 40 degree**Then ^AXB= 2 x 40 = 80. Therefore ^AXB = 80 degree ii**ANSWER IN BRIEF**• How can the angle subtended by an arc at the centre be defined? • The angle in semicircle is ______ degree. • The angle subtended by an arc at the centre is ______ the angle subtended by it at any point on the circle. • Which is the theorem that defines the relationship between the angles subtended by an arc at a centre and at a point on the circle ? P.S. For answers view the slideshow again