Construction of an Equilateral Triangle
Steps of construction
Step I: Draw a ray AX with initial point A.
Step II: With centre A and radius equal to length of a side of the triangle draw an arc BY, cutting the ray AX at B.
Step III: With centre B and the same radius draw an arc cutting the arc BY at C.
Step IV: Join AC and BC to obtain the required triangle.
Read More:
- Construction Of Similar Triangle As Per Given Scale Factor
- Construction Of A Line Segment
- Construction Of The Bisector Of A Given Angle
- Construction Of Perpendicular Bisector Of A Line Segment
- Construction Of An Angle Using Compass And Ruler
Construction of a Triangle when its Base, sum of the other Two Sides and One Base Angle are given
Example 1: Construct a triangle ABC in which AB = 5.8cm, BC + CA = 8.4 cm and ∠B = 60º.
Solution:
Steps of Construction
Step I: Draw AB = 5.8 cm
Step II: Draw ∠ABX = 60º
Step III: From point B, on ray BX, cut off line segment
BD = BC + CA = 8.4 cm.
Step IV: Join AD
Step V: Draw the perpendicular bisector of AD meeting BD at C.
Step VI: Join AC to obtain the required triangle ABC.
Example 2: Construct a triangle ABC, in which BC = 3.8cm, ∠B = 45º and AB + AC = 6.8 cm.
Solution:
Steps of Construction
Step I: Draw BC = 3.8 cm.
Step II: Draw ∠CBX = 45º
Step III: Form B on ray BX, cut-off line segment BD equal to AB + AC i.e. 6.8 cm.
Step IV: Join CD.
Step V: Draw the perpendicular bisector of CD meeting BD at A.
Step VI: Join CA to obtain the required triangle ABC.
Construction of a Triangle when its Base, difference of the other Two Sides and One Base Angle are given
Case (1): ∠A = 30º, AC – BC = 2.5
Case (2): ∠A = 30º, BC – AC = 2.5
Example 1: Construct a triangle ABC in which base
AB = 5 cm, ∠A = 30º and AC – BC = 2.5 cm.
Solution:
Steps of Construction
Step I: Draw base AB = 5 cm
Step II: Draw ∠BAX = 30º
Step III: From point A, on ray AX, cut off line segment
AD = 2.5 cm (= AC – BC).
Step IV: Join BD.
Step V: Draw the perpendicular bisector of BD which cuts AX at C.
Step VI: Join BC to obtain the required triangle ABC.
Example 2: Construct a triangle ABC in which BC = 5.7 cm, ∠B = 45º, AB – AC = 3 cm.
Solution:
Steps of Construction
Step I: Draw base BC = 5.7 cm.
Step II: Draw ∠CBX = 45º
Step III: From B, on ray BX, cut off line segment
BD = 3 cm (= AB – AC).
Step IV: Join CD.
Step V: Draw the perpendicular bisector of CD which cuts BX at A.
Step VI: Join CA to obtain the required triangle ABC.
Example 3: Construct a ∆ABC in which BC = 5.6 cm,
AC – AB = 1.6 cm and ∠B = 45º. Justify your construction.
Solution:
Steps of construction
Step I: Draw BC = 5.6 cm
Step II: At B, construct ∠CBX = 45º
Step III: Produce XB to X’ to form line XBX’.
Step IV: From ray BX’, cut-off line segment BD = 1.6 cm
Step V: Join CD
Step VI: Draw perpendicular bisector of CD which cuts BX at A
Step VII: Join CA to obtain required triangle BAC.
Justification: Since A lies on the perpendicular bisector of CD. Then
∴ AC = AD = AB + DB = AB + 1.6
Hence, ∆ABC is the required triangle.
Construction of a Triangle of given Perimeter and Two Base Angles
Example 1: Construct a triangle PQR whose perimeter is equal to 14 cm, ∠P = 45º and ∠Q = 60º.
Solution:
Steps of Construction
Step I: Draw a line segment XY = 14 cm
Step II: Construct ∠YXD = ∠P = 45º and
∠XYE = ∠Q = 60º
Step III: Draw the bisectors of angles ∠YXD and ∠XYE mark their point of intersection as R.
Step IV: Draw right bisectors of RX and RY meeting XY at P and Q respectively.
Step V: Join PR and QR to obtain the required triangle PQR.
