Answer :
Let the given right-angled triangle be ΔABC.
Steps of Construction:
(i) Draw base, BC = 3.5 cm
Now, draw a ray BX such that ∠XBC = ∠B = 90°.
(ii) Here, sum of two sides = AB + AC = 5.5 cm
So, cut the line segment BD = 5.5 cm from ray BX.
(iii) Join DC.
(iv)Draw perpendicular bisector of DC, say PQ, which intersects BD at A.
(v) Now join AC.
Thus, we get a ΔABC which is the required triangle.
Justification:
Base BC and ∠B are drawn as given.
Since, PQ is perpendicular bisector of CD and A lies on it.
∴ AD = AC
Now, AB = BD – AD
= BD – AC
∴ AB + AC = BD
Thus, construction is justified.
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