Question

in a circle with centre O AB and CD are two diameters perpendicular to each other the length of chord AC is:​

Answer 1

Resposta:

${AB} ^2 + {DC} ^2 ={AC}^2$

Explicação passo-a-passo:

According to the Phitagorian Theorem, in an rectangle triangle the biggest side of the triangle squared is equal to the sum of the other two sides squared.

In this situation we have an rectangle triangle, that has the sides AB CD and AC, being this last one the biggest one, so using the Phitagorian Theorem, the answer is:

${AB} ^2 + {DC} ^2 = {AC} ^2$

Answer 2

Resposta:

AC =  square ruth(2R^2).

Explicação passo-a-passo:

So, you've the the chord AC as the hypotenuse and the chords AB/2 and CD/2, both equal to the radius R. Thus applies Pitagoras, which give us

R^2 + R^2 = AC^2

2R^2 = AC^2

Therefore,

AC =  square ruth(2R^2).