Question

Question 5
Construct an equilateral triangle, given its side and justify the construction.

Answer 1

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 ${60}^{\circ }$.

Steps to construct an equilateral triangle of side 5 cm.

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 5 cm, which intersects extended line segment AE at C.

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


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,
$\angle B=\angle C\left(isoscelesangles\right)$
$\angle A+\angle B+\angle C={180}^{\circ }\left(anglesumpropertyofatriangle\right)$
${60}^{\circ }+\angle C+\angle C={180}^{\circ }{60}^{\circ }+2\angle C={180}^{\circ }2\angle C={180}^{\circ }-{60}^{\circ }={120}^{\circ }\angle C={60}^{\circ }\angle B=\angle C={60}^{\circ }Wehave,\angle A=\angle B=\angle C={60}^{\circ }\dots \dots \left(1\right)\angle A=\angle Band\angle A=\angle CBC=ACandBC=AB\left(sidesoppositetoequalanglesofatriangle\right)AB=BC=AC=5cm\dots \dots \left(2\right)Fromequations\left(1\right)and\left(2\right),\Delta ABCisanequilateraltriangle$.