# Constructions

Notes of chapter: Constructions are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 9.  Step 2-

Now, draw two arcs from points E and D with radius more than ED. Step 3-

Both arcs will intersect at point F. Join F and B.

BF is required bisector of angle ABC.   (ii) To construct the perpendicular bisector of a given line segment.

Given – Line segment AB.

What to construct – Perpendicular bisector of the line segment AB.

Steps of Construction –

Step 1-

Draw a line segment AB. Step 2 –

Draw two arcs from points A and B as centres with radius more than half of the line segment AB.

Step 3-

Draw arcs both sides of the line segment AB Step 4 –

These arcs intersect each other at points C and D.

Step 5 –

Join C and D. CD intersects AB at point E.

CD is our required perpendicular bisector of line segment AB. Justification –

Join A to C and D.

Join B to C and D.   (iii) To construct an angle of 600 at the initial point of a given ray.

Given – A ray AB with initial point A.

What to construct – Draw an angle of 600 at initial point A.

Steps of Construction –

Sep 1-

Draw a ray AB with A as initial point. Step 2-

Draw an arc of any radius which intersect ray AB at point D. Step 3-

Draw an arc with same radius from point D as centre which intersect previous arc at point E.   Justification –

Join E and D.    Step 2-

Cut a line segment BD of measure AC + BC from the ray EA. Step 3-

Join DB. Step 4 –

Draw an angle DBF equal to angle ADB. Step 5 –

Ray BF intersects ray AE at point C.

Therefore, ABC is a triangle.   Step 2-

Draw an angle EAB. Step 3-

Cut a line segment AD equal to AC – BC from AE. Step 4-

Join D to B. Step 5-

Draw perpendicular bisector PQ of DB. Step 6-

Let PQ intersect AE at point C.

Therefore, ABC is required triangle. Justification –

CD = BC (C lies on the perpendicular bisector of DB)

= AC – BC

Hence, proved.  Step 2-

Draw an angle EAB.

Step 3-

Cut a line AD segment BC – AC from AE. Step 4 –

Join D to B. Step 5 –

Draw perpendicular bisector PQ of DB. Step 6 –

Let PQ intersect AE at point C.

Therefore, ABC is required triangle. Justification –

CD = BC (C lies on the perpendicular bisector of DB)

= BC – AC

Hence, proved.

(ii) To construct a triangle, given its perimeter and its two base angles.

Given – Perimeter of triangle ABC = AB + BC + CA,

Two base angles B and C.

What to construct – A triangle ABC.

Steps of Construction –

Step 1 –

Draw a line segment XY equal to perimeter AB + BC + CA. Step 2-

Draw angle LXY equal to B and MYX equal to C. Step 3-

Draw bisectors of angle LXY and angle MYX. Let these bisectors intersect at point A. Step 4-

Draw perpendicular bisector PQ of AX and RS of AY. Step 5-

Let PQ intersect XY at point B and RS intersect XY at point C. Step 6 –

Join AB and AC.  Helping Topics

NCERT Solutions Class 9

Worksheet Class 9

Practical Geometry