Question

Find the perimeter of an isosceles right triangle with each of its congruent sides is $7cm$.

• A. $7(4+\sqrt{2})$
• B. $7(2+\sqrt{2})$
• C. $7(7+\sqrt{2})$
• D. $7(3+\sqrt{2})$

Answer 1

The correct option is B $7(2+\sqrt{2})$

Given the congruent side = 7 cm or Perpendicular (P) = Base (B) = 7 cm
Since, the triangle is a right angled isosceles triangle,
$H^2=B^2+P^2$
$H^2=7^2+7^2$
$H=7\sqrt{2}$ cm
Perimeter = P + B + H
= $7+7+7\sqrt{2}$
= $14+7\sqrt{2}$
= $7(2+\sqrt{2})$