1. AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is :
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By: anil on 05 May 2019 01.57 am
Given : AB = 10 , CD = 24 and EF = 17 cm To find : OB = OD = $$r$$ = ? Solution : A line perpendicular to the chord from the centre of the circle bisects the chord. => $$AF = FB = frac{AB}{2} = frac{10}{2} = 5$$ Similarly, $$CE = ED = 12$$ Let OF = $$x$$ => OE = $$(17-x)$$ In right $$ riangle$$OFB => $$(OB)^2 = (OF)^2 + (FB)^2$$ => $$r^2 = x^2 + 25$$ Now, in right $$ riangle$$OED => $$(OD)^2 = (OE)^2 + (ED)^2$$ => $$r^2 = (17-x)^2 + 144$$ => $$x^2 + 25 = x^2 - 34x + 289 + 144$$ => $$34x = 408$$ => $$x = frac{408}{34} = 12$$ => $$r^2 = 12^2 + 25$$ => $$r = sqrt{169} = 13$$ cm
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