The radius of the circle that is perpendicular to a chord intersects the chord at its midpoint, so it is said to bisect the chord.
If radius of a circle intersects a chord then it bisects the chord only if radius is perpendicular to the chord.
true, because both distances of the chord are congruent
The relationship between the chord and the radius of the circle is Length of the chord = 2r sin(c/2) where r = radius of the circle and c = angle subtended at the center by the chord
False
Radius
Then the radius bisects the chord.
Bisects
Perpendicular.
If radius of a circle intersects a chord then it bisects the chord only if radius is perpendicular to the chord.
Bisects that chord
False
A Chord. Or another radius!
YesAt a right angle
its false
true, because both distances of the chord are congruent
The perpendicular bisector of ANY chord of the circle goes through the center. Each side of a triangle mentioned would be a chord of the circle therefore it is true that the perpendicular bisectors of each side intersect at the center.
Imagine if you will a circle with a chord drawn through it and a line running from the center of that chord to the center of the circle. That line is necessarily perpendicular to the chord. This means you have a right triangle whose hypotenuse is the radius of the circle. The radius is thus given by: r = sqrt{(1/2 chord length)^2 + (length of perpendicular line)^2} The actual formula to find the radius is as follows: r= C squared/8a + a/2, where C is the chord length, and a is the distance from center point of the chord to the circle , and a and C form an angle of 90 degrees. the entire formula before simplification is r = sqrt {(1/2 C)^2 + (r-a)^2}