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CAT Equal Chords and Their Distances from the Centre - (Part 1) - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

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  • 6 Questions around this concept.

Solve by difficulty

Equal chords of a circle are equidistant from the centre. (True/False)

In the adjoining figure, O is center of the circle. AB and CD are chord of circle having length each 10 cm. If OL = 8 cm, then OM = 

 

If two chords of a circle are equally inclined to the diameter through their point of intersection, then the chords are equal. (True/False)

Concepts Covered - 1

Equal Chords and Their Distances from the Centre - (Part 1)

Distance: The length of the perpendicular from a point to a line is the distance of the line from the point.

Let AB be a line and P be a point. Since there are infinite numbers of points on a line, if you join these points to P, you will get infinitely many line segments PL1 , PL2 , PM, PL3 , PL4 , etc  So, which of these is the distance of AB from P? The least length PM to be the distance of AB from P as PM is perpendicular to AB..

Theorem 6 : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Let a given circle C(O,r) wjth chord AB = CD, OL isperpendicular to AB and OM is perpendicular to CD

We need to prove that OL = OM

As we know that the perpendicular from the centre of a circle to a chord bisects the chord

\\\therefore\mathrm{ \quad A L=\frac{1}{2} A B \quad \text { and } \quad C M=\frac{1}{2} C D}\\ \\\because\quad \mathrm{A B=C D} \\\\\Rightarrow \quad\mathrm{\frac{1}{2} A B=\frac{1}{2} C D}\\\\ \Rightarrow \quad \mathrm{A L=C M\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(i)}

Now, in the right triangle OLA and OMC, we have

               AL = CM              (from (i))

              OA = OC              (radius of circle)

       ∠ OLA = ∠ OMC = 90°

Therefore,   ∆ OLA ≅ ∆ OMC    ([by RHS-congruence)

This gives  OL = OM     (Corresponding parts of congruent triangles)

Hence, chord AB and CD are equidistant from O.

The converse of Theorem 6.

Theorem 7 : Chords of circle (or of congruent circles) which are equidistant from the centre (or centers) of a circle are equal in length.

Let's prove this

Let AB and CD are two chords of a circle C(O,r), OL is perpendicualr to AB and OM is perpendicualr to CD such that OL = OM.

We need to prove AB = CD

Join OA and OC

We know that the perpendicular from the centre of a circle to a chord bisects the chord

\\\therefore\mathrm{ \quad A L=\frac{1}{2} A B \quad \text { and } \quad C M=\frac{1}{2} C D}\\

Now, in the right triangle OLA and OMC, we have

              OA = OC              (radius of circle)

              OL = OM              (given)

        ∠ OLA = ∠ OMC = 90°

Therefore,   ∆ OLA ≅ ∆ OMC    ([by RHS-congruence)

Therefore,   AL = CM  or   2AL = 2CM

Hence, AB = CD             \left [\mathrm{\because \quad A L=\frac{1}{2} A B \quad \text { and } \quad C M=\frac{1}{2} C D} \right ]\\

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