CONSTRUCTION OF PERPENDICULAR BISECTOR OF A LINE SEGMENT
About "Construction of perpendicular bisector of a line segment"
Construction of perpendicular bisector of a line segment :
Even though students know what is perpendicular bisector, many students do not know, how to construct perpendicular bisector of a line segment.
Here we are going to see "How to construct perpendicular bisector of a line segment step by step"
Construction of perpendicular bisector of a line segment
To construct a perpendicular bisector of a line segment, you must need the following instruments.
1. Ruler
2. Compass
The steps for the construction of a perpendicular bisector of a line segment are:
Step 1 :
Draw the line segment AB.
Step 2 :
With the two end points A and B of the line segment as centers and more than half the length of the line segment as radius draw arcs to intersect on both sides of the line segment at C and D.
Step 3 :
Join C and D to get the perpendicular bisector of the given line segment AB.
In the above figure, CD is the perpendicular bisector of the line segment AB.
This construction clearly shows how to draw the perpendicular bisector of a given line segment with compass and straightedge or ruler. This bisects the line segment (That is, dividing it into two equal parts) and also perpendicular to it. Since this bisects, it finds the midpoint of the given line segment.
Key Concept - Perpendicular Bisector
The line drawn perpendicular through the midpoint of a given line segment is called the perpendicular bisector of the line segment.
Perpendicular bisector in a triangle
The perpendicular bisector of a triangle is a line which is passing through the mid point of the side and also perpendicular to that side.
Step 1 :
Draw the triangle ABC.
Step 2 :
Select one the sides of the triangle, say AC
With the two end points A and C of the side AC as centers and more than half the length of the side AC as radius draw arcs to intersect on both sides of the side AC and join the points of intersection of the arcs.
Using the steps explained above, in the above triangle ABC, perpendicular bisector is drawn to the side AC.
Similarly we can draw perpendicular bisectors to the sides AB and BC as given below.
Hence, every triangle will have three perpendicular bisectors.
Key Concept - Circumcenter
The point of concurrency of the perpendicular bisectors of the three sides of a triangle is called the circumcenter and is usually denoted by S.
After having gone through the stuff given above, we hope that the students would have understood "Special line segments of triangles"
If you want to know more about "Special line segments of triangles", please click here
If you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Word problems on quadratic equations
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Markup and markdown word problems
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Word problems on sets and venn diagrams
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
