Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
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Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

KnowledgeBoat · Accepted Answer

Let there be a circle with center O and radius r. AB and CD are two chords of this circle. OM ON where OM ⊥ AB and ON ⊥ CD. To prove: AB > CD We know that, The perpendicular from the centre of a circle to a chord bisects the chord. ∴ AM = and CN = Join OA and OC. In right angle triangle OAM, ⇒ OA² = AM² + OM² ⇒ AM² = OA² - OM² ..... (1) In right angle triangle ONC, ⇒ OC² = CN² + ON² ⇒ CN² = OC² - ON² ..... (2) OM ON ⇒ OM² ON² ⇒ -OM² -ON² ⇒ OA² - OM² OC² - ON² [∵ OA = OC] ⇒ AM² CN² [From (1) and (2)] ⇒ ⇒ AB² CD² ⇒ AB² CD² ⇒ AB CD Hence proved that, of any two chords of a circle, the greater chord is nearer to the centre.

Mathematics

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Circles

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In the given figure, C and D are points on the semi-circle described on AB as diameter.
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Mathematics

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Circles

Related Questions

ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.

In the given figure, C and D are points on the semi-circle described on AB as diameter.Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.

In the given figure, C and D are points on the semi-circle described on AB as diameter.
Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.