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 PQ and draw a ray DE ⊥ PQ. 3.Cut the line segment AD of length 3.2 cm from DE. 4.Make angles equal to 30° at A on both sides of AD say ∠CAD and ∠BAD, where B and C lie on PQ. 5.Cut the line segment DC from PQ such that DC = BD Join AC Thus, A ABC is the required triangle. Justification 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.