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

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

Quote:

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

$\displaystyle \displaystyle \begin{align*} r^2 &= r^2 - 2hr + h^2 + a^2 \\ 0 &= -2hr + h^2 + a^2 \\ 2hr - h^2 &= a^2 \end{align*}$

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

Marvellous,

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

Regards,

Clotsworth