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We are given that the height of the triangle ABC = Diagonal of the square with side 2
units.
From the square, PR = h = 2√2 units (using Pythagoras theorem)
So, In triangle ABC, h = 2√2 units
By using Pythagoras theorem,
AC^2 = AB^2 + BC^2= (2√2))^2+ 1^2
AC^2 = 8 + 1 = 9
AC = 3 units.
So, the perimeter of the triangle ABC = 3 + 1 + 2√2 = 4 + 2√2 = 6.82 = 7 units (rounded off to nearest integer)
Ans. (7)