Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.
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Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.

KnowledgeBoat · Accepted Answer

From figure, AB = 1.4 cm and CD = 4 cm OF AB and OE CD. We know that, Perpendicular from the center to the chord, bisects it. ∴ AF = = 0.7 cm. CE = = 2 cm. EF = 3.9 cm and OA = OC = r cm. Let OE = x and OF = 3.9 - x In right-angled triangle OCE, By pythagoras theorem, ⇒ Hypotenuse2 = Perpendicular2 + Base2 ⇒ OC2 = OE2 + CE2 ⇒ OC2 = x2 + 22 ⇒ OC2 = x2 + 4 .....(1) In right-angled triangle OAF, By pythagoras theorem, ⇒ Hypotenuse2 = Perpendicular2 + Base2 ⇒ OA2 = OF2 + AF2 ⇒ OA2 = (3.9 - x)2 + O.72 ⇒ OA2 = (3.9 - x)2 + 0.49 ....(2) From (1) and (2), we get : ⇒ x2 + 4 = (3.9 - x)2 + 0.49 ⇒ x2 + 4 = x2 - 7.8x + 15.21 + 0.49 ⇒ x2 + 4 = x2 - 7.8x + 15.7 ⇒ 4 = 15.7 - 7.8x ⇒ 7.8x = 15.7 - 4 ⇒ 7.8x = 11.7 ⇒ x = ⇒ x = 1.5 Substituting value of x in equation (1), we get : ⇒ OC2 = (1.5)2 + 4 ⇒ OC2 = 2.25 + 4 ⇒ OC2 = 6.25 ⇒ OC = = 2.5 cm. Hence, radius of the circle = 2.5 cm.

Mathematics

Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.

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