Given one side of the equilateral triangle be 2.5cm.
STEPS :
(i) Draw a line segment BC = 2.5cm.
(ii) Through B, construct ray BP making angle 600 with BC.
i.e. ∠PBC= 600
(iii) Through C, construct CQ making angle 600 with BC
i.e., ∠QCB = 600
(iv) Let BP and CQ intersect each other at point A. The n, ΔABC is the require equilateral triangle.
Proof : Since, ∠ABC= ∠ACB = 600
∴ ∠BAC = 1800 - (600 + 600) = 600
=> All the angles of the ΔABC drawn are equal. => All the sides of the ΔABC drawn are equal.
=> ΔABC is the required equilateral triangle. Hence Proved.
Alternate method:
If one side is 2.5cm, then each side of the required equilateral triangle is 2.5cm.
STEPS :
(i) Draw BC = 2.5cm
(ii) With B as centre, draw an arc of radius 2.5cm
(iii) With C as centre, draw an arc of radius 2.5cm
(iv) Let the two arc intersect each other at point A. Join AB and AC. Then, ABC is the required equilateral triangle.
Case (ii) When the base of the triangle, one base angle and the sum of other two sides are given.
