In the same circle or in congruent circles
- Chords equidistant from the center of a circle are congruent
- Congruent chords are equidistant from the center
- The perpendicular bisector of a chord contains the center of the circle
AB =10
Problem 2
Step 1
Step 2
x = 4
We can use the good old pythagorean theorem.
52 = 32 + x2x = 4
Problem 3
Step 1
Step 2
x2 + 72 = 252
x = 24
Chord = 2
× 24 = 48
x = 24
Chord = 2
× 24 = 48
Problem 4
Step 1
Step 2
142 + 482 = r2
r = 50
r = 50
Problem 5
No, not necessarilly. Although the one chord is bisected we do not kow that the two chords are equidistant from the center.