Haryana State Board HBSE 9th Class Maths Notes Chapter 11 Constructions Notes.

## Haryana Board 9th Class Maths Notes Chapter 11 Constructions

**Introduction**

In previous chapters, we have drawn rough diagrams to prove theorems or solving exercise. But sometimes we need to construct accurate figures. For example, to draw a map of building to be constructed, to design the parts of machine and tools etc. To draw such figures the following geometrical instruments are needed:

(i) A graduated scale, (ii) a pair of set squares, (iii) a pair of dividers, (iv) a pair of compasses, (v) a protractor (see in the figure below).

These instruments are used in drawing a geometrical figure, such as a triangle, quadrilateral, trapezium, circle etc. In this chapter we will learn about some simple and basic constructions such as bisector of a line segment, bisector of a given angle, constructions of some standard angle and some constructions of triangles with the help of graduated scale, compass and protractor.

**Key Words**

→ Geometrical Construction: A geometrical construction is the process of drawing a geometrical figure using only two instruments-an ungraduated ruler and compass.

→ Perpendicular bisector: A line which divides the given line in two equal parts and is perpendicular to the line segment is called the perpendicular bisector.

→ Corresponding: Angles, lines and points in one figure which bear a similar relationship, each to each, to angles, lines and points in another figure.

→ Triangle: A closed figure which bounded by three line segments is called a triangle.

→ Isosceles triangle: If at two sides of a triangle are equal, it is called an isosceles triangle.

→ Equilateral triangle: If all three sides of a triangle are equal, it is called an equilateral triangle.

→ Perimeter of a triangle: Sum of the all sides of a triangle is called the perimeter of a triangle.

→ Median: The median of a triangle, corresponding to any side, is the line joining the midpoint of that side with the opposite vertex.

→ Altitude: An altitude of a triangle, corresponding to any side, is the length of perpendicular drawn from the opposite vertex to that side.

**Basic Concepts**

Basic Constructions

(a) Construction 11.1: To construct the bisector of a given angle.

We have an ∠ABC, we need to construct its bisector.

Steps of construction:

Step – I: Taking B as centre and any suitable radius draw an arc cutting AB and BC at P and Q respectively.

Step – II: Taking P and Q as centres and radius more than \(\frac{1}{2}\)PQ, draw two arcs intersecting each other at R.

Step-III: Draw the ray BR. The ray BR is the required bisector of the angle ABC.

Justification: Join PR and QR.

In ΔBPR and ΔBQR, we have

PB = QB [Radii of the same arc]

PR = QR [Arcs of equal radii]

BR = BR [Common]

ΔBPR ≅ ΔBQR [By SSS congruence rule]

⇒ ∠PBR = ∠QBR [CPCT]

Hence, BR is the bisector of ∠ABC.

Construction 11.2:

To construct the perpendicular bisector of a given line segment.

We have a line segment AB, we need to construct its perpendicular bisector.

Steps of construction:

Step-I: Draw a line segment AB of given length.

Step-II: Taking A as centre and radius more than \(\frac{1}{2}\)AB, draw arcs one on each side of AB.

Step-III: Taking B as centre and the same radius as in Step (II), draw arcs intersecting previous arcs at P and Q.

Step-IV: Join PQ intersecting AB at M. Then line PQ is the perpendicular bisector of AB.

Justification: Join AP, BP, AQ and BQ.

In ΔPAQ and ΔPBQ, we have

AP = BP Arcs of equal radii]

AQ = BQ [Arcs of equal radii]

PQ = PQ [Common]

∴ ΔPAQ ≅ ΔPBQ [By SSS congruence rule]

⇒ ∠APQ ≅ ∠BPQ [CPCT]

⇒ ∠APM = ∠BPM ……(i)

In ΔPMA and ΔPMB, we have

PA = PB [Arcs of equal radii]

∠APM = ∠BPM [As proved in (i)]

PM = PM [Common]

ΔPMA ≅ ΔPMB (By SAS congruence rule)

⇒ AM = BM

and ∠PMA = ∠PMB (CPCT)

But, ∠PMA + ∠PMB = 180° [By linear pair axiom]

∠PMA + ∠PMA = 180°

2∠PMA = 180°

∠PMA = \(\frac{180^{\circ}}{2}\) = 90°

∠PMA = ∠PMB = 90°.

Hence, PM is the perpendicular bisector of AB.

Construction 11.3:

To construct an angle of 60° at the initial point of a given ray.

Let us take a ray AB with initial point A.

We need to construct a ray AC such that

∠CAB = 60°.

Steps of construction:

Step-I: Draw a ray AB with initial point A.

Step-II: Taking A as centre and suitable radius, draw an arc which intersects AB at P.

Step-III: Taking P as centre and same radius as before, draw an arc intersecting the previous are at point Q.

Step-IV: Draw a ray AC passing through Q. Then ∠CAB is the required angle of 60°.

Justification: Join PQ.

AP = PQ = AQ [By construction]

⇒ ΔPAQ is an equilateral triangle.

∴ ∠QAP = 60°

⇒ ∠CAB = 60°.

Some Constructions of Triangles:

Construction 11.4 To construct a triangle, given its base, a base angle and sum of other two sides.

Given: The base BC, a base angle, say ∠B and the sum AB + AC of the other two sides of a triangle ABC, we need to construct ΔABC.

Steps of construction:

Step-I: Draw the given base BC and the point B construct an angle ∠CBX equal to the given angle with BC.

Step-II: From the ray BX,

cut BD = AB + AC.

Step-III: Join DC and construct

∠DCY = ∠BDC.

Step-IV: Let CY intersects BX at A.

Then, ABC is the required triangle.

Justification: Since,

∠BDC = ∠DCY (By construction)

⇒ ∠ADC = ∠ACD

⇒ AD = AC

[Sides opp. to equal angles are equal]

Now, AB = BD – AD

⇒ AB = BD – AC [∵AD = AC]

⇒ AB + AC = BD

⇒ BD = AB + AC.

Alternative method

Steps of construction:

Step-I: Draw the given base BC and at the point B construct ∠CBX equal to the given angle with BC.

Step-II: From the ray BX, cut BD = AB + AC.

Step-III: Join DC and draw perpendicular bisector PY of CD intersecting BD at point A. Join AC.

Remarks: The construction of the trinagle is not possible if the sum AB + AC ≤ BC.

Construction 11.5: To construct a triangle given its base, a base angle and the difference of the other two sides.

Given: Base BC and a base angle, say ∠B and the difference of other two sides AB – AC or AC – AB. We need to construct AABC. There are following two cases arise:

Case I: If AB > AC that is AB – AC is given.

Steps of construction:

Step-I: Draw the given base BC. At the point B, construct ∠CBX equal to the given angle with BC.

Step-II: From ray BX, cut the line segment BD is equal to AB – AC.

Step-III Join DC and draw perpendicular bisector of CD intersecting BX at A.

Step-IV: Join AC. Then ABC is required triangle.

Justification: Since AP is the perpendicular to CD.

∴ AD = AC

So, BD = AB – AD

⇒ BD = AB – AC.

Case II: If AB < AC that is AC – AB is given.

Steps of construction:

Step-I: Draw the given base BC. At the point B, construct ∠CBX equal to the given angle with BC.

Step-II: Extend XB to D to the opposite side of BC such that

BD = AC – AB.

Step-III: Join CD and draw perpendicular bisector PQ of CD intersecting BX at A.

Step-IV: Join AC, then ABC is required triangle.

Justification: Since PQ is the perpendicular bisector of CD.

∴ AD = AC

Now, BD = AD – AB

⇒ BD = AC – AB.

Construction 11.6:

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

Given: Base angles ∠B and ∠C and perimeter of a triangle ABC (ie., BC + CA + AB) are given. We need to construct the triangle ABC.

Steps of construction:

Step – I: Draw a line segment XY equal to BC + CA + AB.

Step – II: Construct ∠LXY equal to ∠B and ∠MYX equal to ∠C with XY.

Step – III: Draw the bisectors of ∠LXY and ∠MYX intersecting at A.

Step – IV: Draw perpendicular bisectors PQ of AX and RS of AY.

Step – V: Let perpendicular bisectors PQ and RS intersect XY at B and C respectively.

Step – VI: Join AB and AC, then ABC is the required triangle.

Justification: Since PQ is the perpendicular bisector of AX.

∴ XB = AB …..(i)

Similarly, RS is the perpendicular bisector of AY.

∴ CY = AC ……(ii)

Now, XY = XB + BC + CY

XY = AB + BC + AC [Using (i) & (ii)]

Since, XB = AB

⇒ ∠AXB = ∠XAB ……(iii)

and CY = AC

⇒ ∠CAY = ∠CYA …..(iv)

∠ABC = ∠AXB + ∠XAB

[∵ Exterior angle is equal to its opp. two interior angles]

⇒ ∠ABC = ∠AXB + ∠AXB [Using (iii)]

⇒ ∠ABC = 2∠AXB

⇒ ∠ABC = ∠LXY

Similarly, ∠ACB = ∠MYX.