Thread: proof of relation of radius, chord, chord hight of a circle

Hello, I am ok with this 'till the last step, re-arranging for a;

r^2 = (OS)^2 + a^2 (by Principle of Pathagoras, OS being that part of the radius below a chord and a being half the chord length, h the chord hight)
= (r-h)^2 + a^2
= r^2 - 2hr + h^2 + +a^2


therefor a^2 = 2hr - h^2

can someone explain the method involved in the last step?

Thanks,

Clotsworth

Originally Posted by Clotsworth

Hello, I am ok with this 'till the last step, re-arranging for a;

r^2 = (OS)^2 + a^2 (by Principle of Pathagoras, OS being that part of the radius below a chord and a being half the chord length, h the chord hight)
= (r-h)^2 + a^2
= r^2 - 2hr + h^2 + +a^2


therefor a^2 = 2hr - h^2

can someone explain the method involved in the last step?

Thanks,

Clotsworth

Marvellous,
That is most helpful, and much appreciated, thank you.

Regards,

Clotsworth