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What else can I help you with?
What triangle must always have at least one angle bisector that is also a perpendicular bisector?
any isosceles triangle
In a circle the perpendicular bisector of a chord must pass through the center of the circle?
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
What must you do to construct the midpoint of a segment?
With a straight-edge and a compass:Swing arcs from each end of the segment with the compass (without changing the settings)Connect the intersections of these arcs.The resultant is a perpendicular bisector of the segment.
To construct a perpendicular bisector to a given line segment one must construct two what?
Can't be sure what you're asking, but it would be two circles with equal radii longer that half the length of the segment, using the endpoints as the origins of the two circles.
In the diagram below rs is the perpendicular bisector of pq. statements must be true?
Since rs is the perpendicular bisector of pq, it follows that point s is the midpoint of segment pq, meaning that ps is equal to qs. Additionally, because rs is perpendicular to pq, the angles formed at the intersection (∠prs and ∠qrs) are both right angles (90 degrees). Consequently, any point on line rs is equidistant from points p and q.
When constructing a perpendicular bisector why must the compass opening be greater than the length of the segment?
So that the arc is mid-way in perpendicular to the line segment
To construct a perpendicular bisector to a given line segment one must construct two?
Equilateral triangles
Which types of triangles must always have at least one angle bisector that is also a perpendicular bisector?
iscoceles triangle! =)
What triangle must always have at least one angle bisector that is also a perpendicular bisector?
any isosceles triangle
What must be true about a perpendicular bisector and the segment it bisects?
It bisects the line segment at midpoint at 90 degrees and its slope is the reciprocal of the line segment's slope plus or minus.
In a circle the perpendicular bisector of a chord must pass through the center of the circle?
Yes, in a circle, the perpendicular bisector of a chord does indeed pass through the center of the circle. This is because the perpendicular bisector of a chord divides it into two equal segments and is equidistant from the endpoints of the chord. Since the center of the circle is the point that is equidistant from all points on the circle, it must lie on the perpendicular bisector. Thus, any chord's perpendicular bisector will always intersect the center of the circle.
What must you do to construct the midpoint of a segment?
With a straight-edge and a compass:Swing arcs from each end of the segment with the compass (without changing the settings)Connect the intersections of these arcs.The resultant is a perpendicular bisector of the segment.
How is line m is the perpendicular bisector of XY. If line m intersects XY at point Z statements must be true?
The perpendicular bisector of the line XY will meet it at its midpoint at right angles.
To construct a perpendicular bisector to a given line segment one must construct two what?
Can't be sure what you're asking, but it would be two circles with equal radii longer that half the length of the segment, using the endpoints as the origins of the two circles.
In the diagram below rs is the perpendicular bisector of pq. statements must be true?
Since rs is the perpendicular bisector of pq, it follows that point s is the midpoint of segment pq, meaning that ps is equal to qs. Additionally, because rs is perpendicular to pq, the angles formed at the intersection (∠prs and ∠qrs) are both right angles (90 degrees). Consequently, any point on line rs is equidistant from points p and q.
What Line t is the perpendicular bisector of FG. If line t intersects FG at point H statements must be true Check all that apply.?
The answer letters always rearrange so here are the answers point H is the midpoint of FG line t intersects FG at a right angle Line T is perpendicular to FG
Line BE is the bisector of segment AC. If AB 7 then AC .?
If line BE is the bisector of segment AC, it means that BE divides AC into two equal segments. Therefore, if AB is 7, then AC must be twice that length, making AC equal to 14.
