Question

Find the perimeter of an isosceles right triangle with each of its congruent as 7cm.

• A. $7\sqrt{2}$ cm
• B. $14$ cm
• C. $(2+\sqrt{2})$ cm
• D. $7(2+\sqrt{2})$ cm

Answer 1

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

Let the other side of triangle is x cm

Then in isosceles right angle triangle two congruent sides are 7 cm

$x^2=(7{)}^2+(7{)}^2$

$\Rightarrow x^2=49+49$

$\Rightarrow x^2=198$

$\Rightarrow x=7\sqrt{2}$

Then perimeter of right angle isosceles triangle =$7+7+7\sqrt{2}=14+7\sqrt{2}=7(2+\sqrt{2})$

So, option D is correct.