Let given right triangle be ABC. Then, given BC = 3.5 cm, ∠B = 90° and sum of other side and hypotenuse i.e., AB + AC = 5.5 cm To construct ΔABC use the following steps 1.Draw the base BC = 3.5 cm 2.Make an angle XBC = 90° at the point B of base BC. 3.Cut the line segment BD equal to AB + AC i.e., 5.5 cm from the ray XB. 4.Join DC and make an ∠DCY equal to ∠BDC. 5.Let Y intersect BX at A. Thus, ΔABC is the required triangle. Justification Base BC and ∠B are drawn as given. In ΔACD, ∠ACD = ∠ADC [by construction] AD = AC …(i) [sides opposite to equal angles are equal] Now, AB = BD – AD = BD – AC [from Eq. (i)] => BD = AB + AC Thus, our construction is justified.