Diameter of a circle is 26 cm and length of a chord of the circle is 2

1.	Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR ⊥ chord PQ, P-R-Q, PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR)
Answer» Radius (OP) =\(\frac{d}{2}\) = \(\frac{26}{2}\)= 13 cm ……(i) ∴ PR = 1/2 PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.] = \(\frac{1}{2}\) x 24 = 12 cm …..(ii) In ∆ORP, ∠ORP = 90° ∴ OP² = OR² + PR² [Pythagoras theorem] ∴ 13² = OR² + 12² [From (i) and (ii)] ∴ 169 = OR² + 144 ∴ OR² = 169 – 144 ∴ OR² = 25 ∴ OR = √25 = 5 cm [Taking square root on both sides] ∴ The distance of the chord from the centre of the circle is 5 cm.

Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR ⊥ chord PQ, P-R-Q, PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR)

Answer»

Radius (OP) =\(\frac{d}{2}\) = \(\frac{26}{2}\)= 13 cm ……(i)

∴ PR = 1/2 PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]

= \(\frac{1}{2}\) x 24 = 12 cm …..(ii)

In ∆ORP, ∠ORP = 90°

∴ OP² = OR² + PR² [Pythagoras theorem]

∴ 13² = OR² + 12² [From (i) and (ii)]

∴ 169 = OR² + 144

∴ OR² = 169 – 144

∴ OR² = 25

∴ OR = √25 = 5 cm [Taking square root on both sides]

∴ The distance of the chord from the centre of the circle is 5 cm.

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