An equilateral triangle,

An equilateral triangle, if its altitude is 3.2 cm.


We know that, in an equilateral triangle all sides are equal and all angles are equal i.e., each angle is of 60°.

Given, altitude of an equilateral triangle say ABC is 3.2 cm. To construct the ΔABC use the following steps.

1. Draw a line PQ.

2. Take a point $D$ on $P Q$ and draw a ray $D E \perp P Q$.

3. Cut the line segment $A D$ of length $3.2 \mathrm{~cm}$ from $D E$.

4. Make angles equal to $30^{\circ}$ at $A$ on both sides of $A D$ say $\angle C A D$ and $\angle B A D$, where $B$ and $C$ lie on $P Q$.

5. Cut the line segment $D C$ from $P Q$ such that $D C=B D$1. Draw a line PQ.


Join AC


Thus, $A A B C$ is the required triangle.


Here, ∠A = ∠BAD + ∠CAD

= 30°+30° =60°.

Also, AD ⊥ SC

∴ ∠ADS = 90°.

In ΔABD, ∠BAD + ∠DBA = 180° [angle sum property]

30° + 90° + ∠DBA = 180° [∠BAD = 30°, by construction ]

∠DBA = 60°

Similary, ∠DCA = 60°

Thus, ∠A = ∠B=∠C = 60°

Hence, ΔABC is an equilateral triangle.


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