in a circle of radius 25cm,two parallel chords of length 14cm and 48 cm respectively are drawn on the same side of the centre .find the distance between them.

Dear Student!

Given AB = 14 cm, CD = 48 cm and radius of the circle = 25 cm.

∴ OD = OB = 25 cm

OP ⊥ AB

 (Perpendicular from the centre of the circle to a chord bisects the chord)

OP ⊥ CD,

(Perpendicular from the centre of the circle to a chord bisects the chord)

In right ΔOQB,

OB2 = OQ2 + QB2 (Pythagoras Theorem)

∴ OQ2 = OB2 + QB2

⇒ OQ2 = (25 cm)2 – (7 cm)2

⇒ OQ2 = 625 cm2 – 49 cm2  = 576 cm2

In right ΔOPD,

OD2 = OP2 + PD2 (Pythagoras Theorem)

∴ OP2 = OD2 – PD2

⇒ OP2 = (25 cm)2 – (24 cm)2

⇒ OP2 = 625 cm2 – 576 cm2  = 49 cm2

Distance between the chords = PQ = OQ – QP = 24 cm – 7 cm = 17 cm

Thus, the distance between the chords is 17 cm.

Cheers!

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thank u sir

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