NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.6 – Part 2

Angle subtended by the segment at center is twice the angle subtended by the same segment at cicumference

Angle Subtended by the Segment at Center and at Cicumference

Angle subtended by the segment at center is twice the angle subtended by the same segment at cicumference

Q-3 The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Solution:

Circle c Circle c: Circle through B_1 with center O Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [R, S] Segment h Segment h: Segment [O, A] Segment i Segment i: Segment [P, O] Segment j Segment j: Segment [R, O] Point O O = (0.98, 3.06) Point O O = (0.98, 3.06) Point O O = (0.98, 3.06) Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c Point A Point A: Point on f Point A Point A: Point on f Point A Point A: Point on f Point B Point B: Intersection point of g, h Point B Point B: Intersection point of g, h Point B Point B: Intersection point of g, h

Points P, Q, R, and S Are Concyclic

Points P, Q, R, S, on circle, also chords PQ and RS, AO⊥PQ and OB⊥RS, PQ∥RS

Given,

  • PQ and RS are two parallel chords of a circle with centre O

  • PQ=6cm , RS=8cm and OA=4cm

  • The radius of the circle is rcm .

Since AOPQ and OBRS and PQRS , therefore O, Q, and P are collinear.

Since perpendiculars from center bisect the chord, therefore

  • PA=AQ=12PQ=12(6)=3cm

  • RB=BS=12RS=12(8)=4cm

OAP

  • OP2=OA2+AP2

  • r2=(4)2+(3)2

  • r2=16+9

  • r2=25

  • r=5

ORB

  • OR2=RB2+BO2

  • r2=(4)2+BO2

  • 52=16+BO2

  • 25=16+BO2

  • BO2=9

  • BO=3

So, the distance of chord RS from the centre is 3cm ( RSOB )

Q-4 Let the vertex of an angle QPR be located outside a circle and let the sides of the angle intersect equal chords QA and SR with the circle. Prove that QPR is equal to half the difference of the angles subtended by the chords QR and AS at the centre.

Solution:

Circle c Circle c: Circle through R with center O Segment f Segment f: Segment [R, P] Segment f Segment f: Segment [R, P] Segment g Segment g: Segment [P, Q] Segment g Segment g: Segment [P, Q] Segment h Segment h: Segment [Q, R] Segment i Segment i: Segment [Q, O] Segment j Segment j: Segment [A, S] Segment k Segment k: Segment [O, A] Segment l Segment l: Segment [O, S] Segment m Segment m: Segment [O, R] Segment n Segment n: Segment [A, R] Point O O = (0.46, 2.28) Point O O = (0.46, 2.28) Point O O = (0.46, 2.28) Point R R = (2.24, 0.78) Point R R = (2.24, 0.78) Point R R = (2.24, 0.78) Point P P = (-3.18, 0.78) Point P P = (-3.18, 0.78) Point P P = (-3.18, 0.78) Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point A Point A: Intersection point of c, g Point A Point A: Intersection point of c, g Point A Point A: Intersection point of c, g Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c

PQR Is a Triangle With Vertex P Located Outside a Circle

PQR is a triangle with vertex P located outside a circle, the sides of the angle intersect equal chords OA and RS.

In a triangle sum of external angle is equal to the sum of interior opposite angles. Therefore, in PAR , we have

  • QAR=APR+ARP …………equation (1)

Also angle subtended at the centre is twice the angle at any point on the remaining part of circle, therefore,

  • QAR=12QOR

  • ARP=12AOS ………equation (2)

From (1) and (2), we have

  • 12QOR=QPR+12AOS ( APR=QPR )

  • QPR=12(QORAOS)

So, QPR is equal to half the difference of angles subtended by the chords QR and AS at the centre.

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