In a circle PQ & RS are two chords bisecting each other, prove that the two parts of one chord are equal to the two parts of the other.

see proof below

Step-by-step explanation:

Let

p1,p2 = half lengths of chord p

q1,q2 = half length of chord q

By the intersecting chord theorem,

p1*p2 = q1*q2,    substituting p1=p2, q1=q2

p1^2 = q1*2

Take square-roots and reject negative roots

p1 = q1

therefore

p1=p2 = q1=q2, or

two parts of one chord are equal to the two parts of the other.

its 4 because 2 times 2 is 4, if you double the number two you get four

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6