NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.5 – Part 3

Opposite angles of a cyclic quadrilateral

Opposite angles of a cyclic quadrilateral, a+b=180

Opposite Angles of a Cyclic Quadrilateral Are Supplementry

Opposite angles of a cyclic quadrilateral, a+b=180

Q-6 PQRS is a cyclic quadrilateral whose diagonals intersect at a point O. If SQR=70° , QPRis30° , find QRS . Further, if PQ=RS , find ORS .

Solution:

Circle c Circle c: Circle through B with center A Angle ? Angle ?: Angle between S, Q, C' Angle ? Angle ?: Angle between S, Q, C' Angle ? Angle ?: Angle between S, Q, C' Angle ? Angle ?: Angle between R, P, D' Angle ? Angle ?: Angle between R, P, D' Angle ? Angle ?: Angle between R, P, D' Segment f Segment f: Segment [S, R] Segment g Segment g: Segment [S, Q] Segment h Segment h: Segment [Q, R] Segment i Segment i: Segment [Q, P] Segment j Segment j: Segment [P, R] Segment k Segment k: Segment [P, S] Point S Point S: Point on c Point S Point S: Point on c Point S Point S: 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 Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point O Point O: Intersection point of g, j Point O Point O: Intersection point of g, j Point O Point O: Intersection point of g, j

P, Q, R, S Are Points of on Circle

P, Q, R, S are points of on circle. It center is O, SR is and PQ are chords

In the figure on circle four points are P, Q, R and S at center O

For chord SR,

  • RQS=RPS (Angles in same segment)

  • RPS=70°

  • QPR=30°

  • QPS=QPR+RPS=30°+70°=100°

  • QRS+QPS=180° (Opposite angles of a cyclic quadrilateral)

  • QRS+100°=180°

  • QRS=80°

In ΔPQR

  • PQ=QR (given)

  • QRP=RPQ (Angles opposite to equal sides of a triangle)

  • QRP=30° ( QPR=30° )

Also, QRS=80°

  • QRP+PRS=80°

  • 30°+PRS=80°

  • PRS=50°

ORS=50° (Angle PRS and ORS are same angles)

Q-7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral; prove that it is a rectangle.

Solution:

Circle c Circle c: Circle through B with center O Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [Q, R] Segment h Segment h: Segment [R, S] Segment i Segment i: Segment [S, P] Segment j Segment j: Segment [Q, S] Segment k Segment k: Segment [P, R] Point O O = (-1.22, 3.24) Point O O = (-1.22, 3.24) Point O O = (-1.22, 3.24) 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

PORS Is Quadrilateral on Circle.

PORS is quadrilateral on circle.PR and SQ are the diameters of the circle

Given PQRS is cyclic quadrilateral inside the circle, and its diagonals, PR and SQ are the diameters of the circle,

Note that chords of equal length subtend the same angle on the circle, in this case PR and SQ are diameters, therefore the angles they subtend on the circle will be equal and also 90° . That is, PQR=QRS=RSP=SPQ=90° (Angles in the semi-circle). Thus, PQRS is a rectangle as each internal angle is 90°.

Q-8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

Circle c Circle c: Circle through B with center A Angle ? Angle ?: Angle between R, N, F' Angle ? Angle ?: Angle between R, N, F' Angle ? Angle ?: Angle between R, N, F' Angle ? Angle ?: Angle between N, M, H' Angle ? Angle ?: Angle between N, M, H' Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [S, R] Segment h Segment h: Segment [P, M] Segment i Segment i: Segment [Q, N] Segment j Segment j: Segment [P, S] Segment k Segment k: Segment [Q, R] 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 S Point S: Point on c Point S Point S: Point on c Point S Point S: 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 M Point M: Point on g Point M Point M: Point on g Point M Point M: Point on g Point N Point N: Point on g Point N Point N: Point on g Point N Point N: Point on g

PQRS Is a Trapezium on Circle

PQRS is a trapezium on circle, PS and QR are equal, PM and QN are perpendicular

Given that, PQRS is the trapezium with, PS and QR pair of opposite equal sides.

Construct perpendiculars PM and QN on SR from vertices P and Q respectively.

Now, in ΔPSM and ΔQRN ,

  • PS=QR (Given)

  • SMP=RNQ (Right angles)

  • PM=QN (Distance between the parallel lines)

Therefore, ΔPSMΔQRN by Right Angle Hypotenuse congruence and , S=R by Corresponding Parts of Congruent Triangles

Also,

  • R+Q=180° (sum of the co-interior angles of parallel lines)

  • S+Q=180° (S=R )

Thus, PQRS is a cyclic quadrilateral as the pair of opposite angles are supplementary.

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