NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.5 – Part 4

Vertically opposite anges formed by two intersecting lines

Vertically Opposite Angles

Vertically opposite anges formed by two intersecting lines

Vertical opposite angles are the angles that are vertically opposite to each other when two lines intersect

Q-9 Two circles intersect at two points Q and B. Through Q, two line segments PQA and RQS are drawn to intersect the circles at P, A and R, and S respectively (see Fig.). Prove that PBR=SBA.

Circle c Circle c: Circle through B_1 with center A_1 Circle d Circle d: Circle through D with center C Segment f Segment f: Segment [R, P] Segment g Segment g: Segment [R, S] Segment h Segment h: Segment [A, P] Segment i Segment i: Segment [P, B] Segment j Segment j: Segment [A, B] Segment k Segment k: Segment [R, B] Segment l Segment l: Segment [S, B] Segment m Segment m: Segment [A, S] Point P Point P: Point on c Point P Point P: Point on c Point P Point P: 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 B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d Point B Point B: Intersection point of c, d Point Q Point Q: Intersection point of c, d Point Q Point Q: Intersection point of c, d Point Q Point Q: Intersection point of c, d Point A Point A: Point on d Point A Point A: Point on d Point A Point A: Point on d Point S Point S: Point on d Point S Point S: Point on d Point S Point S: Point on d

Two Circles Intersecting at Two Points Q and B.

Two circles intersect at two points Q and B, two line segments PQA and RQS are drawn to intersect the circle at P, A and R, and S respectively.

Solution:

  • Here, two circles intersect at two point Q and B.

  • Two line segment PQA and RQS are drawn to intersect the circles at A,D and P,Q respectively

  • Now, join PR and AS chords.

    For chord PR,

    • RQP=PBR (Angles in the same segment) ………equation (1)

    For chord AS,

  • AQS=SBA (Angles in same segment) …………equation (2)

  • PQA and RQS are line segments intersecting at Q, therefore RQP=AQS (Vertically opposite angles)………….equation (3)

By the equations (1)

  • RQP=PBR

  • AQS=PBR ( RQP=AQS From equation (3))

  • PBR=SBA ( AQS=SBA From equation (2))

Q-10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

Circle c Circle c: Circle through B with center A_1 Circle d Circle d: Circle through D with center C Segment f Segment f: Segment [Q, R] Segment g Segment g: Segment [Q, P] Segment h Segment h: Segment [P, R] Segment i Segment i: Segment [P, S] 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 d Point R Point R: Point on d Point R Point R: Point on d Point P Point P: Intersection point of c, d Point P Point P: Intersection point of c, d Point P Point P: Intersection point of c, d Point S Point S: Intersection point of c, d Point S Point S: Intersection point of c, d Point S Point S: Intersection point of c, d

Two Circles With Diameters on Triangle

Two circles are the side PQ and PR of the triangle ΔPQR as diameter.

  • Given, two circles are drawn on the sides PQ and PR of the triangle ΔPQR as diameters.

  • The circles intersect at P and S.

To prove, S is on QR. We have to prove that QSR is a straight line, that is, PSQ&PSR are supplementary.

PSQ=PSR=90° (Angle in the semi circle)

Therefore, PSQ+PSR=180° and QSR is straight line.

Thus, S on the QR.

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