Yes because 2 of its 3 sides meet at right angles
Perpendicular bisector lines intersect at right angles
That will depend on what type of triangle it is as for example if it is an isosceles triangle then it will form two congruent right angle triangles.
A circle cannot form a perpendicular bisector.
A triangle has no parallel sides but in the form of a right angle triangle it has perpendicular lines that meet at right angles which is 90 degrees.
It has a pair of perpendicular lines that meet at right angles to form the 90 degree angle of a right angle triangle.
Perpendicular bisector lines intersect at right angles
That will depend on what type of triangle it is as for example if it is an isosceles triangle then it will form two congruent right angle triangles.
A circle cannot form a perpendicular bisector.
A circle cannot form a perpendicular bisector.
A circle itself does not form a perpendicular bisector because a perpendicular bisector is a line that divides a segment into two equal parts at a right angle, typically associated with straight segments. However, the concept of a perpendicular bisector can be applied to chords within a circle. The perpendicular bisector of a chord will always pass through the center of the circle.
Yes a right angle triangle has perpendicular lines that form 90 degrees.
A triangle has no parallel sides but in the form of a right angle triangle it has perpendicular lines that meet at right angles which is 90 degrees.
Only when it is in the form of a right angle triangle that it will have perpendicular lines that meet at 90 degrees.
A perpendicular bisector intersects a line segment at a right angle, forming two 90-degree angles with the segment. This means that the angle between the bisector and the line segment is always a right angle, indicating that the bisector divides the segment into two equal parts.
It has a pair of perpendicular lines that meet at right angles to form the 90 degree angle of a right angle triangle.
No, they cannot.
Perpendicular lines intersect at right angles.