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1. AB and AC are two chords of a circle of radius'r'Distance of chord AB and chord AC from centre are p' and qrespectively 2AC = AB then prove: 4q² = p²+ 3r² |
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Answer» ANSWER : Given: AB and AC are two chords of a circle with center O. Such that AB=2AC p and q are ⊥ distances of AB and AC from center O i.e., OM=p and ON=q r is the radius of the circle To prove that: =p 2 +3r 2
Proof: Join OA. OM and ON are ⊥ distances of AB and AC from center O. Here, AN= 2 AC
(perpendicular from center to chord intersect at mid-point of the chord) AM= 2 AB
(perpendicular from center to chord intersect at mid-point of the chord) In right angled ΔOMA, OM 2 +AM 2 =OA 2
p 2 +AM 2 =r 2
AM 2 =r 2 −p 2 …… (1) In right angled ΔONA, ON 2 +AN 2 =OA 2
q 2 +AN 2 =r 2
AN 2 =r 2 −q 2 …… (2) Since, AM= 2 AB
= 2 2AC
=AC=2AN From equations (1) and (2), we have r 2 −p 2 =AM 2
r 2 −p 2 =4AN 2
r 2 −p 2 =4[r 2 −q 2 ] 4q 2 =p 2 +3r 2
LHS=RHS |
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