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Mathematics

Assertion (A): The line segment joining the mid-points of all parallel chords of a circle passes through the centre.

Reason (R): If the chords are on same side of centre then only the line through the mid-points of the chords passes through the centre.

  1. A is true, R is false.
  2. A is false, R is true.
  3. Both A and R are true.
  4. Both A and R are false.

Circles

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Answer

A is true, R is false.

Explanation

The line segment joining the mid-points of all parallel chords of a circle passes through the centre. Assertion Reasoning, Concise Mathematics Solutions ICSE Class 9.

Let AB and CD be two parallel chords of a circle with center O. P is mid-point of AB and Q is mid-point of CD.

We have to prove that the line joining the points P and Q passes through the center O, i.e., ∠ POQ = 180°.

Construction : Join OP and OQ. Also, draw OE parallel to AB and CD.

Proof : Since, line segment joining the mid-point of the chord with the center of the circle is perpendicular to the chord, therefore, OP is perpendicular to AB and OQ is perpendicular to CD.

⇒ ∠ OPA = 90° and ∠ OQC = 90°

Now, OE ∥ AB and OP is transversal.

∴ ∠ POE = ∠ OPA = 90° (Alternate angles)

Similarly, OE ∥ CD and OQ is transversal.

∴ ∠ QOE = ∠ OQC = 90° (Alternate angles)

∴ ∠ POQ = ∠ POE + ∠ QOE

= 90° + 90°

= 180°

⇒ POQ is a straight line.

i.e, the line joining the mid-points of two parallel chords passes through the center.

∴ Assertion (A) is true.

If the chords are on same side of centre then only the line through the mid-points of the chords passes through the centre. Assertion Reasoning, Concise Mathematics Solutions ICSE Class 9.

Let AB and CD be two parallel chords that are lying on the same side of a circle with center O. M is mid-point of AB and N is mid-point of CD.

We have to prove that the line joining the points M and N passes through the center O, i.e., ∠ MNO = 180°.

Construction : Join OM and ON. Also, draw OE parallel to AB and CD.

Proof : Since, line segment joining the mid-point of the chord with the center of the circle is perpendicular to the chord, therefore, OM is perpendicular to AB and ON is perpendicular to CD.

⇒ ∠ OMA = 90° and ∠ ONC = 90°

Now, OE ∥ AB and OM is transversal.

∴ ∠ MOE = ∠ OMA = 90° (Alternate angles)

Similarly, OE ∥ CD and ON is transversal.

∴ ∠ NOE = ∠ ONC = 90° (Alternate angles)

∴ ∠ MNO = ∠ MND + ∠ DNO

= ∠ MNC + ∠ CNO

= 90° + 90°

= 180°

⇒ MNO is a straight line.

i.e, The line passing through M and N also passes through O.

∴ Reason (R) is false.

Hence, Assertion (A) is true, Reason (R) is false.

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