Perpendicular Bisector of Triangle: Meaning, Construction & Examples
Perpendicular Bisectorof a Triangle: A perpendicular bisector is a line that cuts a line segment into two equal parts. It typically creates or forms an angle of \(90°\) with the bisected line. The perpendicular bisectors of the sides of a triangle are concurrent, i.e., they meet at one point.
Two lines are perpendicular to each other when they intersect to form \(90°\) with each other. Furthermore, a bisector divides a line into two halves. Thus, a perpendicular bisector of a line segment \({PQ}\) implies that it intersects \({PQ}\) at \(90°\) and cuts it into two halves. Every point in the perpendicular bisector is equidistant from point \({P}\) and \({Q}\).
Definition of Perpendicular Bisector of Triangle
A perpendicular bisector of a triangle is a line passing through the midpoint of each side perpendicular to the given side.
The three perpendicular bisectors of the sides of a triangle meet at one point, called the circumcentre. A point where three or more lines intersect is called a point of coincidence. So, the circumcentre is the point of coincidence of perpendicular bisectors of a triangle. It is equidistant from the vertices of a triangle. That is \(AO=BO=CO\).
What is a Perpendicular Bisector?
A line segment that intersects another line segment at a right angle and it divides that another line into two equal parts at its midpoint is called a perpendicular bisector.
In the above figure, \(CD\) is the perpendicular bisector of \(XY\).
Properties of a Perpendicular Bisector
Perpendicular bisector divides a line segment into two halves or bisects it.
Perpendicular bisector makes right angles with (or is perpendicular to) a line segment.
Every point in the perpendicular bisector is equidistant from both the ends of the line segment.
Bisection of a Line segment
Bisection of a line segment \(PQ\) means dividing it into two equal parts or finding a point \(O\) on \(PQ\) such that \(PO = OQ\).
One way of bisecting a line segment is to measure it and mark off a point exactly half the length from one end of the line segment. However, this method of bisecting lines is not accurate because the divisions on a ruler are limited. Therefore, an accurate measurement of the line segment is not possible.
This section shall learn the procedure of bisecting a given line segment using a ruler and compasses only. The below mentioned steps are used to do the same.
Steps to Draw a Perpendicular Bisector
How to draw a perpendicular bisector?
Draw a line segment \(PQ\) of the given length.
With centre \(P\) and radius more than half of \(PQ\), draw arcs, one on each side of \(PQ\).
With \(Q\) as the centre and the same radius as before, draw arcs, cutting the previously drawn arcs at \(R\) and \(S\) respectively.
Join \(RS\) intersecting \(PQ\) at \(O\). Then \(O\) bisects the line segment \(PQ\) as shown below. If we measure \(PO\) and \(OQ\), it will be equal \( \Rightarrow PO = OQ\).
Construction of Perpendicular Bisectors of Triangle
A perpendicular bisector of a triangle is a line passing through the midpoint of each side perpendicular to the given side.
How to draw a perpendicular bisector of a triangle?
Below are the steps to construct the perpendicular bisector of a triangle.
Construct a triangle \(\Delta ABC\).
With the help of a compass placed at vertex \(B\), take the length equal to more than half of \(AB\) and draw the arcs on both sides.
From the vertex \(A\), with the same length cut the arcs which drawn from vertex \(B\).
Draw a line passing through the intersection, and this line is the perpendicular bisector of side \(AB\).
To draw a perpendicular bisector of side \(AC\), take a compass, place it on vertex \(C\), take the length equal to more than half of \(AC\) and draw the arcs on both sides.
From the vertex \(A\), with the same length cut the arcs which drawn from vertex \(C\).
Draw a line passing through the intersection, and this line is the perpendicular bisector of side \(AC\).
To draw a perpendicular bisector of side \(BC\), take a compass, place it on vertex \(B\), take the length equal to more than half of \(BC\) and draw the arcs on both sides.
From the vertex \(C\), with the same length cut the arcs which drawn from vertex \(B\).
Draw a line passing through the intersection, and this line is the perpendicular bisector of side \(BC\).
The point of intersection \(I\) of three perpendicular bisectors is called a circumcentre.
Properties of Perpendicular Bisector of Sides of a Triangle
A triangle has three perpendicular bisectors.
The perpendicular bisectors of a triangle are concurrent.
The point of concurrence is called a circumcentre.
In an acute angle triangle, the circumcentre is in the interior of the triangle.
In a right-angled triangle, the hypotenuse passes through the circumcentre.
In an obtuse angle triangle, the circumcentre is on the outside of the triangle.
Practice Problems for Perpendicular Bisector of Triangle – Solved Examples
Q.1.Construct a triangle \(ABC\), \(∠B = 60°\), \(∠ C = 45°\) and \(AB + BC + CA = 11\,\rm{cm}\). Ans: Steps of construction:
Draw a line segment \(XY\) equal to \(AB + BC + CA = 11\,\rm{cm}\)
Make the angle equal to \(∠ B = 60°\) from point \(X\). Let the angle be \(∠ LXY\).
Make the angle equal to \(∠ C = 45°\) from point \(Y\). Let the angle be \(∠ MYX\).
Bisect angles \(∠ LXY\) and \(∠ MYX\). Let these bisectors intersect at point \(A\).
Make the perpendicular bisector of \(AX\). Let it intersect \(XY\) at point \(B\).
Make the perpendicular bisector of \(AY\). Let it intersect \(XY\) at point \(C\).
Join \(AB\) and \(AC\). Therefore, \(△ ABC\) is the required triangle.
Q.2. Construct a \(△ PQR\) in which \(QR = 4.2\,\rm{cm}\), \(∠ Q = 120°\) and \(PQ = 3.5\,\rm{cm}\). Draw \(PM ⊥ QR\). Ans: Steps of construction:
Draw a line segment \(QR = 4.2\,\rm{cm}\)..
Construct \(∠ RQX = 120°\). Mark point \(P\) at a distance of \(3.5\,\rm{cm}\) from \(Q\).
Join \(PR\). Then, \(△ PQR\) is the required triangle.
Produce \(RQ\) to \(S\) to the left.
With \(P\) as a centre and with a different radius, draw an arc, cutting \(SQ\) at \(A\) and \(B\).
With \(A\) as centre and radius more than half of \(AB\), draw an arc below \(SR\). Now, with \(B\) as the centre and with the same radius, draw another arc, cutting the previous arc at \(C\).
Join \(PC\), meeting \(RQ\) produced at \(M\). Then, \(PM ⊥ QR\).
Q.3. Construct a \(△ ABC\) in which \(BC = 3.6\,\rm{cm}\), \(AB = 5\,\rm{cm}\) and \(AC = 5.4\,\rm{cm}\). Draw the perpendicular bisector of the side \(BC\). Ans: Steps of construction:
Draw a line segment \(AC = 5.4\,\rm{cm}\).
With \(A\) as centre and radius \(5\,\rm{cm}\), draw an arc.
With \(C\) as a centre and radius \(3.6\,\rm{cm}\), draw another arc, cutting the previous arc at \(B\)
Join \(AB\) and \(CB\). Then, \(△ ABC\) is the required triangle.
With \(B\) as centre and radius measuring more than half of \(BC\), draw arcs on both sides of \(BC\).
With \(C\) as the centre and the same radius before, draw arcs on both sides of \(BC\), cutting the previous arcs at \(P\) and \(Q\) as shown. Join \(PQ\).
Therefore, \(PQ\) is the required perpendicular bisector of \(BC\), meeting \(BC\) at \(M\).
Q.4. Construct a triangle \(△ ABC\) in which \(AB = AC = 4.8\,\rm{cm}\) and \(BC = 5.3\,\rm{cm}\). Measure \(∠ B\) and \(∠ C\). Draw \(AD ⊥ BC\). Ans: Steps of construction:
Draw a line segment \(BC = 5.3\,\rm{cm}\)
With \(B\) as centre and radius \(4.8\,\rm{cm}\), draw an arc.
With \(C\) as a centre and radius \(4.8\,\rm{cm}\), draw another arc, cutting the previous arc at \(A\).
Join \(AB\) and \(AC\). Then, \(△ ABC\) is the required triangle.
When we measure \(∠ B\) and \(∠ C\) the angles of the triangle by protractor, then the measure of angles are \(56°\) and \(56°\) respectively.
With \(A\) as the centre and any radius, draw an arc cutting \(BC\) at \(M\) and \(N\).
With \(M\) as the centre and the radius more than half of \(MN\), draw an arc.
With \(N\) as the centre and the same radius, draw another arc cutting the previously drawn.
Join \(AP\) cutting \(BC\) at \(D\). Then, \(AD ⊥ BC\).
Q.5. Construct a triangle \( △ ABC\) in which \(BC = 5.6\,\rm{cm}\), \(AC – AB = 1.6\,\rm{cm}\) and \(∠ B = 45°\). Ans: Steps of construction
Draw \(BC = 5.6\,\rm{cm}\).
At \(B\), construct \(∠ CBX = 45°\).
Produce \(XB\) to \(X^{’}\) to form line \(XBX^{’}\)
From ray \(BX^{’}\), cut- off the line segment \(BD = 1.6\,\rm{cm}\)
Join \(CD\)
Draw perpendicular bisector of \(CD\) which cuts \(BX\) at \(A\)
Join \(CA\) to obtain the required triangle \(△ ABC\).
Summary
In this article, we learned about the definition of the perpendicular bisector of the triangle, the construction of the perpendicular bisector of a triangle and solved some examples on the perpendicular bisector of a triangle. Hope this article will help you understand and solve problems on this topic.
FAQs on Perpendicular Bisector of Triangle
Q.1. How do you find the perpendicular bisector of a triangle? Ans: If a line segment intersects a side of a triangle at a right angle and divides that line into two equal parts at its midpoint, then it is a perpendicular bisector of a triangle.
Q.2. Explain the perpendicular bisector of a triangle? Ans:A perpendicular bisector of a triangle is a line passing through the midpoint of each side perpendicular to the given side.
Q.3. When can you use the perpendicular bisector theorem? Ans: We can use the perpendicular bisector to measure the distance between one end and the perpendicular. If a pillar is standing at the centre of a temple at an angle, all the points on the pillar will be equidistant from the endpoints of the temple.
Q.4. How do you prove two lines are perpendicular? Ans: If two lines are perpendicular to each other, then the angle between the two lines will be \(90°\).
Q.5. What is the perpendicular bisector theorem? Ans: The perpendicular bisector theorem states that any point on the perpendicular bisector is equidistant from both the endpoints of the line segment on which it is drawn.
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Practice Questions and Mock Tests for Maths (Class 8 to 12)
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