Find the perimeter of the isosceles right angled triangle shown in the above figure if the length of its hypotenuse is given to be equal to $8 \sqrt{2}$ .

8
$8 + 16 \sqrt{2}$
16
$8 + 8 \sqrt{2}$
$16 + 8 \sqrt{2}$

Question

Find the perimeter of the isosceles right angled triangle shown in the above figure if the length of its hypotenuse is given to be equal to $8 \sqrt{2}$ .

8
$8 + 16 \sqrt{2}$
16
$8 + 8 \sqrt{2}$
$16 + 8 \sqrt{2}$

Toppr Admin · Accepted Answer

In isosceles right angle triangle the length of the other sides - h-x221A-2Let the sides is x- then-x21D2-x-h-x221A-2-x21D2-x-8-x221A-2-x221A-2-x21D2-x-8Perimeter of isosceles right angled triangle - 2x-h-x21D2-2-xD7-8-8-x221A-2-x21D2-16-8-x221A-2

Find the perimeter of the isosceles right angled triangle shown in the above figure if the length of its hypotenuse is given to be equal to 8√2.88+16√2168+8√216+8√2

In isosceles right angle triangle the length of the other sides = h√2Let the sides is x, then⇒x=h√2⇒x=8√2√2⇒x=8Perimeter of isosceles right angled triangle = 2x+h⇒2×8+8√2⇒16+8√2

Find the perimeter of the isosceles right angled triangle shown in the above figure if the length of its hypotenuse is given to be equal to $8 \sqrt{2}$ .

8
$8 + 16 \sqrt{2}$
16
$8 + 8 \sqrt{2}$
$16 + 8 \sqrt{2}$