Two parallel chords of lengths of 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between chords.
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Question

Two parallel chords of lengths of 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between chords.

KnowledgeBoat · Accepted Answer

Let AB and CD be chords of length 16 cm and 30 cm respectively. From figure, We know that, Perpendicular from the center to the chord, bisects it. ∴ AF = = 8 cm. CE = = 15 cm. From figure, OA = OC = radius = 17 cm. In right-angled triangle OCE, By pythagoras theorem, ⇒ Hypotenuse2 = Perpendicular2 + Base2 ⇒ OC2 = OE2 + CE2 ⇒ 172 = OE2 + 152 ⇒ 289 = OE2 + 225 ⇒ OE2 = 289 - 225 ⇒ OE2 = 64 ⇒ OE = = 8 cm. In right-angled triangle OAF, By pythagoras theorem, ⇒ Hypotenuse2 = Perpendicular2 + Base2 ⇒ OA2 = OF2 + AF2 ⇒ 172 = OF2 + 82 ⇒ 289 = OF2 + 64 ⇒ OF2 = 289 - 64 ⇒ OF2 = 225 ⇒ OF = = 15 cm. From figure, ⇒ EF = OE + OF = 8 + 15 = 23 cm. Hence, distance between the chords = 23 cm.

Mathematics

Two parallel chords of lengths of 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between chords.

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