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Question
AB and AC are the two chords of a circle whose radius is r. If p and q are
the distance of chord AB and CD, from the centre respectively and if
AB = 2AC then proove that 4q2 = p2 + 3r2.
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Solution
Let AC = a then AB = 2a
seg OM ⊥ chord AC, seg ON ⊥ chord AB.
AM =MC = `a/2` and AN = NB = a
In Δ OMA and Δ ONA, By theorem of Pythagoras,
`AO^2 = AM^2 + MO^2`
`AO^2 = AN^2+ q^2` ....... (1)
`AO^2 = AN^2 + NO^2`
`AO^2 = a^2 + p^2` ........ (2)
From equation (1) and (2)
`(a/2)^2 + q^2 = a^2 + p^2`
`a^2/4 + q^2 = a^2 + p^2`
`a^2 + 4q^2 = 4a^2 + 4p^2`
`4q^2 = 3a^2 + 4p^2`
`4q^2 = p^2 + 3(a^2 + p^2)`
`4q^2 = p^2 + 3r^2 .... ("In" Δ ONA, r^2 = a^2 + p^2)`
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