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 ${60}^{\circ }$.

Steps to construct an equilateral triangle of side $5cm.$

Step 1: Draw a line segment $AB$ of length $5cm.$ Draw an arc of some radius, taking A as its centre. Let it intersect $AB$ at $P.$

Step 2: Taking $P$ as the centre, draw an arc to intersect the previous arc at E and join $AE.$

Step 3: Taking $A$ as the centre, draw an arc of radius $5cm,$ which intersects extended line segment $AE$ at $C.$

Step 4: Join $AC$ and $BC,$ $\Delta ABC$ is the required equilateral triangle of side $5cm.$


Justification,
We can justify the construction by showing $△ABC$ 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 of isosceles triangles angles)
$\angle A+\angle B+\angle C={180}^{\circ }$ (angle sum property of a triangle)
${60}^{\circ }+\angle C+\angle C={180}^{\circ }$
${60}^{\circ }+2\angle C={180}^{\circ }$
$2\angle C={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\angle C={60}^{\circ }$
$\angle B=\angle C={60}^{\circ }$
We have,$\angle A=\angle B=\angle C={60}^{\circ }\dots \dots \left(1\right)$
$\angle A=\angle Band\angle A=\angle C$
$BC=ACandBC=AB$ (sides opposite to equal angles of a triangle)
$AB=BC=AC=5cm\dots \dots \left(2\right)$
From equations $\left(1\right)and\left(2\right),$ $\Delta ABC$ is an equilateral triangle.