Constructions Class 9 Notes: Chapter 11

According to the CBSE Syllabus 2023-24, this chapter has been removed from NCERT Class 9 Maths textbook.

CBSE Class 9 Maths Constructions Notes:-Download PDF Here

Introduction to Constructions

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 angles form a straight line.

Angle Bisector

For more information on Angle Bisector, watch the below video

Construction of an Angle bisector

Suppose we want to draw the angle bisector of ∠ABC we will do it as follows:

• Taking B as the centre and any radius, draw an arc to intersect AB and BC to intersect at D and E, respectively.
• Taking D and E as centres and with a radius of more than DE/2, draw arcs to intersect each other at a point F.
• Draw the ray BF. This ray BF is the required bisector of the ∠ABC.

To know more about Angle Bisector, visit here.

Perpendicular Bisector

Construction of a perpendicular bisector

Steps of construction of a perpendicular bisector on the line segment AB:

• Take A and B as centres and radii more than AB/2 draw arcs on both sides of the line.
• Arcs intersect at points C and D. Join CD.
• CD intersects AB at M. CMD is the required perpendicular bisector of the line segment AB.

Proof of the validity of construction of a perpendicular bisector

Proof of the validity of the construction of the perpendicular bisector:
ΔDAC and ΔDBC are congruent by SSS congruency. (∵ AC = BC, AD = BD and CD = CD)
∠ACM and ∠BCMare equal (cpct)
ΔAMC and ΔBMC are congruent by SAS congruency. (∵ AC = BC, ∠ACM = ∠BCM and CM = CM)
AM = BM and ∠AMC = ∠BMC (CPCT)
∠AMC + ∠BMC = 180⁰ (Linear Pair Axiom)
∴ ∠AMC = ∠BMC = 90^∘
Therefore, CMD is the perpendicular bisector.

To know more about Perpendicular Bisector, visit here.

Constructing Angles

Construction of an Angle of 60 Degrees

Steps of construction of an angle of 60 degrees:

• Draw a ray QR.
• Take Q as the centre and some radius draw an arc of a circle, which intersects QR at a point Y.
• Take Y as the centre with the same radius; draw an arc intersecting the previously drawn arc at point X.
• Draw a ray QP passing through X
• ∠PQR=60^∘

Proof for the validity of construction of an angle of 60 degrees

Proof for the validity of construction of the 60⁰ angle:
Join XY
XY = XQ = YQ (By construction)
∴△XQY is an equilateral triangle.
Therefore, ∠XQY = ∠PQR = 60⁰

To know more about Constructing Angles, visit here.

Triangle Constructions

Construction of Triangles

At least three parts of a triangle have to be given for constructing it, but not all combinations of three parts are sufficient for the purpose.
Therefore a unique triangle can be constructed if the  following parts of a triangle are given:

• Two sides and the included angle is given.
• Three sides are given.
• Two angles and the included side is given.
• In a right triangle, a hypotenuse and one side are given.
• If two sides and an angle (not the included angle) are given, it is not always possible to construct such a triangle uniquely.

Given base, base angle and the sum of other two sides

Steps for construction of a triangle given base, base angle, and the sum of the other two sides:

• 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 make an angle DCY equal to ∠BDC.
• Let CY intersect BX at A.
• ABC is the required triangle.

Given base(BC), base angle(ABC) and AB-AC

Steps of construction of a triangle given base(BC), base angle(∠ABC) and difference of the other two sides (AB-AC):

• Draw base BC and with point B as the vertex make an angle XBC equal to the given angle.
• Cut the line segment BD equal to AB – AC(AB > AC) on the ray BX.
• Join DC and draw the perpendicular bisector PQ of DC.
• Let it intersect BX at a point A. Join AC.
• Then △ABC is the required triangle.

Proof for validation for construction of a triangle with given base, base angle and difference between two sides

Validation of the steps of construction of a triangle with given base, base angle and difference between two sides

• Base BC and ∠B are drawn as given.
• Point A lies on the perpendicular bisector of DC. So, AD = AC.
• BD = AB – AD = AB – AC (∵ AD = AC).
• Therefore ABC is the required triangle.

Given base (BC), base angle (ABC) and AC-AB

Steps of construction of a triangle given base (BC), base angle (∠ABC) and difference of the other two sides (AC-AB):

• Draw the base BC and at point B make an angle XBC equal to the given angle.
• Cut the line segment BD equal to AC – AB from the line BX extended on the 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.
• △ABC is the required triangle.

Given perimeter and two base angles

Steps of construction of a triangle with a given perimeter and two base angles.

• Draw a line segment, say GH equal to BC + CA + AB.
• Make angles XGH equal to ∠B and YHG equal to ∠C, where angle B and C are the given base angles.
• Draw the angle bisector of∠XGH and ∠YHG. Let these bisectors intersect at a point A.
• Draw perpendicular bisectors PQ of AG and RS of AH.
• Let PQ intersect GH at B and RS intersect GH at C. Join AB and AC
• △ABC is the required triangle.

To know more about Triangle Constructions, visit here.

Proof for validation for the construction of a triangle with a given perimeter and two base angles

Validating the steps of construction of a triangle with a given perimeter and two base angles:

• B lies on the perpendicular bisector PQ of AG and C lies on the perpendicular bisector RS of AH. So, GB = AB and CH = AC.
• BC + CA + AB = BC + GB + CH = GH (∵ GB = AB and CH = AC)
• ∠BAG = ∠AGB (∵ΔAGB, AB = GB)
• ∠ABC = ∠BAG + ∠AGB = 2∠AGB = ∠XGH
• Similarly, ∠ACB =∠YHG