The perimeter of an isosceles right-angled triangle having a as each of the equal sides is
(a) $(1 + \sqrt{2}) a$
(b) $(2 + \sqrt{2}) a$
(c) $3 a$
(d) $(3 + \sqrt{2}) a$

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Solution

(b) $(2 + \sqrt{2}) a$
For an isosceles right-angled triangle having each equal side as a, we have:
Hypotenuse = $\sqrt{a^{2} + a^{2}} = \sqrt{2 a^{2}} = \sqrt{2} a$
Perimeter = $a + a + \sqrt{2} a = 2 a + \sqrt{2} a = (2 + \sqrt{2}) a$

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