class ix ncert cbse construction chapter part 1

1. CONSTRUCTIONS
N.C.E.RT CHAPTER-11
Part-1

2. Basic Constructions
In earlier classes you might have learnt to construct perpendicular bisector of a line
segment, angles of 30°,45°, 60°, 90°, 120° 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.

3. 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 centre 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.

4. 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.

5. To Construct an angle of 60°
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 60° 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