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

Download PDF of This Page (Size: 161K)

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:

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 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 .

Solution:

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

  • That is is an isosceles triangle

Proof

In cyclic quadileteral PQRA, ………….equation (1) (opposite angles of cyclic quadileteral are supplementary)

We know that RSA is a straight line. Therefore, ………….equation (2)

In a parallelogram opposide angles are equal, are opposite angles of a parallelogram therefore, . Therefore from 2 we can write ………….equation (3)

From equation (1) and (3)

Since in ,we have

  • Therefore,