Constructions

Chapter 11 Class 9th Mathematics chapter 11

To Construct an Angle Equal to a Given Angle

Given: Any POQ and a point A

Required: To construct an angle at A equal to POQ

Steps of Construction:

With O as centre and any (suitable) radius, draw an arc to meet OP at R and OQ at S.

Through A draw a line AB.

Taking A as centre and same radius (as in step 1), draw an arc to meet AB at D.

Measure the segment RS with compasses.

With D as centre and radius equal to RS, draw an arc to meet the previous arc at E.
Join AE and produce it to C, then BAC is the required angle equal to POQ.

Linear Pair axiom

If a ray stands on a line then the adjacent angles form a linear pair of angles.
If two angles form a linear pair, then uncommon arms of both the angles form a straight line.

To Bisect a Given Angle

Given: Any POQ

Required: To bisect POQ.

Steps of Construction:

1. With O as centre and any (suitable) radius, draw an arc to meet OP at R and OQ at S.
2. With R as centre and radius more than half of RS, draw an arc. Also, with S as centre and same radius draw another arc to meet the previous arc at T.

Join OT and produce it, then OT is the required bisector of POQ.

To Construct some Specific Angles

To construct an angle of 60°

Steps of Construction:

Draw any line OP.
With O as centre and any suitable radius, draw an arc to meet OP at R.
With R as centre and same radius (as in step 2), draw an arc to meet the previous arc at S.
Join OS and produce it to Q, then POQ = 60°.

To construct an angle of 30°

Steps of Construction

Construct POQ = 60°.
Bisect POQ. Let OT be the bisector of POQ, then POT = 30°

To construct an angle of 120°

Steps of Construction:

Draw any line OP.
With O as centre and any suitable radius, draw an arc to meet OP at R.
With R as centre and same radius (as in step 2), draw an arc to meet the previous arc at T. With T as centre and same radius, draw another arc to cut the first arc at S.
Join OS and produce it to Q, then POQ = 120°.

To construct an angle of 90°

Steps of Construction:

Construct POQ = 60°
Construct POV = 120°.
Bisect QOV. Let OU be the bisector of QOV, then POU = 90°.

To construct an angle of 45°

Steps of Construction:

Construct AOP = 90°.
Bisect AOP.
Let OQ be the bisector of AOP, then AOQ = 45°

To Draw a Perpendicular Bisector of a Line Segment

Given: Any line segment PQ.

Required: To draw a perpendicular bisector of line segment PQ.

Steps of Construction:

With P as centre, take a length greater than half of PQ and draw arcs one on each side of PQ.
With Q as centre and same radius (as in step 1), draw two arcs on each side of PQ cutting the previous arcs at A and B.
Join AB to meet PQ at M, then AB bisects PQ at M, and is perpendicular to PQ, Thus, AB is the required perpendicular bisector of PQ.

Properties of a Perpendicular Bisector

It divides AB into two equal halves or bisects it.
It makes right angles with (or is perpendicular to) AB.
Every point in the perpendicular bisector is equidistant from point A and B.

While working with practical geometry, you will often find the application of perpendicular bisectors; say when you are asked to draw an isosceles triangle, or when you have to determine the centre of a circle, etc. Below are the steps to construct a perpendicular bisector of a line using a compass and a ruler.

How to Construct a Perpendicular Bisector?

You will require a ruler and compasses. The steps for the construction of a perpendicular bisector of a line segment are:

Step 1: Draw a line segment PQ.

Step 2: Adjust the compass with a length of a little more than half of the length of PQ.

Step 3: Place the compass pointer at point P and draw arcs above and below the line.

Step 4: Keeping the same length in the compass, place the compass pointer at point Q. Similarly, draw two arcs above and below the line keeping the compass pointer at Q.

Step 5: Mark the points where the opposite arcs cross as X and Y.

Step 6: Using a ruler, draw a line passing across X and Y.

The perpendicular bisector bisects PQ at a point J, that is, the length PJ is equal to JQ. And the angle between the two lines is 90 degrees.

Construction of a Triangle, given its Base, sum of the other two sides and one Base Angle

To construct ΔABC in which base BC, B and sum AC + AB of other two sides are given.

Steps of construction:

Draw the base BC and at the point B, make an angle, say XBC equal to the given angle.
Cut a line segment BD = AC + AB from the ray BX.
Join DC and make angle DCY equal to angle BDC.
Let CY intersect BX at A.
ABC is the required triangle.

Alternate Method

Steps of construction:

Draw the base BC and at the point B, make an angle, say XBC equal to the given angle.
Cut a line segment BD = AC + AB from the ray BX.
Draw perpendicular bisector PQ of CD to intersect BD at a point A. Join AC.

ABC is the required triangle.

Construction of a Triangle, given its Base, difference of the other two sides and one Base

To construct ΔABC when the base BC, a base angle B and the difference of other two sides AB – AC or AC – AB are given.

Case 1: When AB > AC and AB – AC is given

Steps of construction:

Draw the base BC and at point B make an angle say XBC equal to the given angle.
Cut the line segment BD equal to AB – AC from ray BX.
Join DC and draw the perpendicular bisector, say PQ of DC. Let it intersect BX at a point A. Join AC Then, ABC is the required triangle.

Case 1: When AB < AC and AC – AB is given Steps of Construction:

Draw the base BC and at point B make an angle say XBC equal to the given angle.
Cut a line segment BD equal to AC – AB from the line BX extended on opposite side of line segment BC.
Join DC and draw the perpendicular bisector, say PQ of DC.
Let PQ intersect BX at A. Join AC.

Then, ABC is the required triangle.