NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.6 – Part 3

Q-5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Solution:

Circle c Circle c: Circle through B_1 with center Q Angle ? Angle ?: Angle between g, h Angle ? Angle ?: Angle between g, h Angle ? Angle ?: Angle between g, h Segment f Segment f: Segment [D, C] Segment g Segment g: Segment [B, D] Segment i Segment i: Segment [D, A] Segment j Segment j: Segment [B, C] Segment k Segment k: Segment [A, B] Segment h Segment h: Segment [A, C] Segment h Segment h: Segment [A, C] Point Q Q = (0.46, 2.5) Point Q Q = (0.46, 2.5) Point Q Q = (0.46, 2.5) Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point B B = (3.25, 5.85) Point B B = (3.25, 5.85) Point B B = (3.25, 5.85) Point A A = (-0.25, 5.83) Point A A = (-0.25, 5.83) Point A A = (-0.25, 5.83) Point E Point E: Point on i Point E Point E: Point on i Point E Point E: Point on i Point F Point F: Point on j Point F Point F: Point on j Point F Point F: Point on j Point P Point P: Point on k Point P Point P: Point on k Point P Point P: Point on k Point O Point O: Intersection point of c, g Point O Point O: Intersection point of c, g Point O Point O: Intersection point of c, g

ABCD Is a Rhombus

ABCD is a rhombus, with diagonals AC and BD intersecting each other at O.

Given,

A rhombus ABCD whose diagonals AC and BD intersect each other at O.

To prove, a circle with CD as diameter passes through O.

Since the diagonals of the rhombus bisect each other at right angles, therefore DOC is right angle

Also note that DC is the diameter of the circle. We know that angle inside the semicirce is a right angle, therefore a right triangle with DC as hypotenues would have its third vertex on the circle. Therefore point O at the intersectiof of two diagonals of the rhombus must be on the circle.

Q-6 PQRS is a parallelogram.The circle through P,Q and R intersect SR (produced if necessary) at A. Prove that PA=PS .

Solution:

Circle c Circle c: Circle through B with center A_1 Segment f Segment f: Segment [A, R] Segment g Segment g: Segment [P, Q] Segment h Segment h: Segment [A, P] Segment i Segment i: Segment [R, Q] Segment j Segment j: Segment [P, S] Point A Point A: Point on c Point A Point A: Point on c Point A Point A: 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 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 f Point S Point S: Point on f Point S Point S: Point on f

a Parallelogram PQRS

PQRS is a parallelogram, with a circle through P,Q and R intersecting RS.

Given,

  • PQRS is a parallelogram

  • The circle through P,Q and R intersects SR at A

To Prove

  • PA=PS

  • That is PAS is an isosceles triangle

  • orPAS=PSA.

Proof

In cyclic quadileteral PQRA, PAS+PQR=180° ………….equation (1) (opposite angles of cyclic quadileteral are supplementary)

We know that RSA is a straight line. Therefore, PSA+PSR=180° ………….equation (2)

In a parallelogram opposide angles are equal, PSRandPQR are opposite angles of a parallelogram therefore, PSR=PQR . Therefore from 2 we can write PSA+PQR=180° ………….equation (3)

From equation (1) and (3)

  • PAS+PQR=PSA+PQR

  • PAS=PSA

Since in PAS ,we have

  • PAS=PSA

  • Therefore, PS=PA

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