Math, asked by Akanksha111, 1 year ago

In the figure, AB and AC are two cords of a circle of centre O and radius r. If AB=2AC and the perpendiculars drawn from the centre on these chords are of lengths 'a' and 'b' , respectively , prove that 4b2 (square)=a2 (square) +3r2(square)..

plzz help... this ques. is making me go crazy...

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Answered by chayannitdgp10

follow the image.it has been solved there..

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Akanksha111: thanx a lot ... I will definitely help you in future..

chayannitdgp10: You are welcome.. Happy to help you..

Akanksha111: exuse me ... i m sry but i need a little more help

chayannitdgp10: Yeah.. How can I help you..??????

Akanksha111: i didnt get the last two lines..

chayannitdgp10: In equation (iv) multiple with 3 on both side.... Then add a^2 on both side... By these we get 4a^2 which is equal to b^2.. replace it... then u will get what u want...

chayannitdgp10: Did you get the last part..??????

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In the figure, AB and AC are two cords of a circle of centre O and radius r. If AB=2AC and the perpendiculars drawn from the centre on these chords are of lengths 'a' and 'b' , respectively , prove that 4b2 (square)=a2 (square) +3r2(square).. plzz help... this ques. is making me go crazy...

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In the figure, AB and AC are two cords of a circle of centre O and radius r. If AB=2AC and the perpendiculars drawn from the centre on these chords are of lengths 'a' and 'b' , respectively , prove that 4b2 (square)=a2 (square) +3r2(square)..

plzz help... this ques. is making me go crazy...