In earlier classes you might have learnt to construct perpendicular bisector of a line segment, angles of 300,450 600, 900, 1200 and also to draw bisector of the given angle. An angle bisector is a ray, which divides an angle in to two equal parts. A line bisector is a line that cuts a line segment into two equal halves. A perpendicular bisector is a line, which divides a given line segment into two equal halves and is also perpendicular to the line segment.
To construct the bisector of a given angle.
Let’s consider angle DEF, we want to construct the bisector of angle DEF.
Steps of construction:
- With E as centre and small radius draw arcs on the rays DE and EF.
- Let the arcs intersect the rays DE and EF at G and H respectively.
- With centres G and H draw two more arcs with the same radius such that they intersect at a point . Let the intersecting point be I
- Now draw a ray with E as the starting point passing through I
- EI is the bisector of the angle DEF.
To construct a perpendicular bisector of a line segment
Lets consider the line segment as PQ. We have to construct the perpendicular bisector of PQ.
Steps of Construction:
- Draw a line segment PQ.
- With P as centre draw two arcs on either sides of PQ with radius more the half the length of the given line segment.
- Similarly draw two more arcs with same radius from point Q such that they intersect the previous arcs at R and S respectively.
- Join the Points R and S. RS is the required perpendicular bisector of the given line segment PQ.
To Construct an angle of 600 at the initial point of a given ray.
Let us take ray PQ with P as the initial point. We have to construct a ray PR such that it makes angle of 600 with PQ.
Steps of Constructions:
- Draw a ray PQ
- With P as centre draw an arc with small radius such that it intersects ray PQ at C.
- With C as centre and same radius draw another arc to intersect the previous arc at D.
- Draw a ray PR from point P through D. Hence the angle RPQ is equal to 60 degrees.
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Constructions of Triangles
You have already learnt how to construct a triangle when three measurements are given. Even we can construct a triangle when the base, one base angle and the sum of the other two sides are given or given its base, a base angle and the difference of the other two sides or given, its perimeter and two base angles.
In
QR = ‘a’cm, 
and PQ + PR = ‘b’ cm.
Step 1: Draw the base QR = ‘a’ cm.
Step 2: Draw
XQR =
.
Step 3: Mark an arc S on QX such that QS = ‘b’ cm.
Step 4: Join RS.
Step 5: Draw the perpendicular bisector of RS such that it intersects QS at P.
Step 6: Join PR.
Thus,
is the required triangle.
In
, given BC = ‘a’ cm, 
and difference of two sides AB and AC is equal to ‘b’ cm.
Case I: AB > AC
Step 1: Draw the base BC = ‘a’ cm.
Step 2: Make
.
Step 3: Mark a point D on ray BX such that BD = ‘b’ cm.
Step 4: Join DC.
Step 5: Draw perpendicular bisector of DC such that, it intersects ray BX at a point A.
Step 6: Join AC.
Thus, ABC is the required triangle.Case II: AB < AC
Step 1: Draw the base BC = ‘a’ cm.
Step 2: Make
and extend ray BX in the opposite direction.
Step 3: Mark a point D on the extended ray BX such that BD = ‘b’cm.
Step 4: Join DC.
Step 5: Draw a perpendicular bisector of DC such that, it intersects ray BX at point A.
Step 6: Join AC.
Thus, ABC is the required triangle.
To construct
, given Perimeter ( AB + BC + CA) = ‘a’ cm, 
.Steps of construction:
Step 1: Draw XY = ‘a’ cm.
Step 2: Draw the ray XL at X making an angle of
with XY.
Step 3: Draw the ray YM at Y making an angle of
with XY.
Step 4: Draw angle bisector of
Step 5: Draw angle bisector of
such that it meets the angle bisector of 
at a point A.
Step 6: Draw the perpendicular bisector of AX such that it meets XY at a point B.
Step 7: Draw the perpendicular bisector of AY such that it meets XY at a point C.
Step 8: Join AB and AC.
Thus, ABC is the required triangle.
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