**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.