Question

Construct an equilateral triangle, given its side and justify the construction

Answer 1

Let us draw an equilateral triangle of side $5cm$. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be $5cm$. We also know that each angle of an equilateral triangle is ${60}^{\circ }$.

The below given steps will be followed to draw an equilateral triangle of $5cm$ side.

Step I: Draw a line segment $AB$ of $5cm$ 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 $5cm$ radius, which intersects extended line segment $AE$ at $C$. Join $AC$ and $BC$. $\Delta ABC$ is the required equilateral triangle of side $5cm$.

Justification of Construction:

We can justify the construction by showing $\Delta ABC$ as an equilateral triangle i.e., $AB=BC=AC=5cm$ and $\angle A=\angle B=\angle C={60}^{\circ }$.

In $\Delta ABC$, we have $AC=AB=5cm$ and $\angle A={60}^{\circ }$.
Since $AC=AB$,
$\angle B=\angle C$ (Angles opposite to equal sides of a triangle)

In $\Delta ABC$,
$\angle A+\angle B+\angle C={180}^{\circ }$ (Angle sum property of a triangle)
$\Rightarrow {60}^{\circ }+\angle C+\angle C={180}^{\circ }$
$\Rightarrow 60°+2\angle C={180}^{\circ }$
$\Rightarrow 2\angle C=180-60={120}^{\circ }$
$\angle C={60}^{\circ }$
$\therefore ,\angle B=\angle C={60}^{\circ }$
We have, $\angle A=\angle B=\angle C={60}^{\circ }$ ... (1)
$\Rightarrow \angle A=\angle B$ and $\angle A=\angle C$
$\Rightarrow BC=AC$and $BC=AB$ (Sides opposite to equal angles of a triangle)
$\Rightarrow AB=BC=AC=5cm$ ... (2)
From equations (1) and (2), $\Delta ABC$ is an equilateral triangle.