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To prove.CD=2OO'
CONSTRUCTION.DRAW OA AND OD PERPENDICULAR TO CD.
NOW, OA PERPENDICULAR CD THAT MEANS OA IS ALSO PERPENDICULAR TO CP.SINCE THE PERPENDICULAR IS DRAWN FROM THE CENTER OF THE CIRCLE TO THE CHORD BISECT THE CHORD .
Therefore,AC IS EQUAL TO AP
CP IS EQUAL TO 2AP
ALSO,
OB IS PERPENDICULAR TO CD
OB IS PERPENDICULAR TO PD
SINCE THE PERPENDICULAR DRAWN FROM THE CENTER OF THE CIRCLE TO THE CHORD BISECT THE CHORD.
THEREFORE,
BP IS EQUAL TO PD
PD IS EQUAL TO 2BP
NOW,
CD IS EQUAL TO 2AP+2BP
CD IS EQUAL TO 2(AP+BP)
CD IS EQUAL TO 2OO'
HENCE,
PROVED.
PLZ MARK ME AS BRAINLIEST.....
CONSTRUCTION.DRAW OA AND OD PERPENDICULAR TO CD.
NOW, OA PERPENDICULAR CD THAT MEANS OA IS ALSO PERPENDICULAR TO CP.SINCE THE PERPENDICULAR IS DRAWN FROM THE CENTER OF THE CIRCLE TO THE CHORD BISECT THE CHORD .
Therefore,AC IS EQUAL TO AP
CP IS EQUAL TO 2AP
ALSO,
OB IS PERPENDICULAR TO CD
OB IS PERPENDICULAR TO PD
SINCE THE PERPENDICULAR DRAWN FROM THE CENTER OF THE CIRCLE TO THE CHORD BISECT THE CHORD.
THEREFORE,
BP IS EQUAL TO PD
PD IS EQUAL TO 2BP
NOW,
CD IS EQUAL TO 2AP+2BP
CD IS EQUAL TO 2(AP+BP)
CD IS EQUAL TO 2OO'
HENCE,
PROVED.
PLZ MARK ME AS BRAINLIEST.....
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