Let us draw an equilateral triangle of side 5 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 5 cm. We also know that each angle of an equilateral triangle is 60º.
The below given steps will be followed to draw an equilateral triangle of 5 cm side.
Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P.
Step II: Taking P as centre, draw an arc to intersect the previous arc at E. Join AE.
Step III: Taking A as centre, draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 5 cm.
Justification of Construction:
We can justify the construction by showing ABC as an equilateral triangle i.e., AB = BC = AC = 5 cm and ∠A = ∠B = ∠C = 60°.
In ΔABC, we have AC = AB = 5 cm and ∠A = 60°.
Since AC = AB,
∠B = ∠C (Angles opposite to equal sides of a triangle)
In ΔABC,
∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)
∠ 60° + ∠C + ∠C = 180°
∠ 60° + 2 ∠C = 180°
2 ∠C = 180° − 60° = 120°
∠C = 60°
∠B = ∠C = 60°
We have, ∠A = ∠B = ∠C = 60° ... (1)
∠A = ∠B and ∠A = ∠C
∠ BC = AC and BC = AB (Sides opposite to equal angles of a triangle)
∠ AB = BC = AC = 5 cm ... (2)
From equations (1) and (2), ΔABC is an equilateral triangle.
