Prove that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point.

Solution:

Consider a circle with diameter AB and centre O

The diameter AB is passing through the point R inside the circle

PQ is a chord perpendicular to AB

CD is another chord passing through R

We have to prove that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point.

Draw OM perpendicular to CD through O.

Consider right angled triangle OMR,

Hypotenuse OR is the longest side of the triangle

OR > OM

OR = Distance of the chord PQ from centre O

OM = Distance of the chord CD from centre O

We know that the chord nearer to the centre is larger than the chord which is farther to the centre.

Therefore, chord PQ is the smallest.

✦ Try This: ABCD is a cyclic quadrilateral in which ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate ∠CAB

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10

NCERT Exemplar Class 9 Maths Exercise 10.4 Sample Problem 2

Prove that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point.

Summary:

It is proven that among all the chords of a circle passing through a given point inside the circle that one is smallest which is perpendicular to the diameter passing through the point

☛ Related Questions:

• If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to . . . .
• If non-parallel sides of a trapezium are equal, prove that it is cyclic
• If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicula . . . .