Construct an equilateral triangle whose sides are 4cm each and justify the construction also.
Please find below the solution to the asked query :
Let us draw an equilateral triangle of side 4 cm. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be 4 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 4 cm side.
Step I: Draw a line segment AB of 4 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 4 cm radius, which intersects extended line segment AE at C. Join AC and BC. ΔABC is the required equilateral triangle of side 4 cm.
Justification of Construction:
We can justify the construction by showing ABC as an equilateral triangle i.e., AB = BC = AC = 4 cm and ∠A = ∠B = ∠C = 60°.
In ΔABC, we have AC = AB = 4 cm and ∠A = 60°.
Since AC = AB,
∠B = ∠C (Angles opposite to equal sides of a triangle)
∠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 = 4 cm ... (2)
From equations (1) and (2), ΔABC is an equilateral triangle.
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