Thread: Geometry question I can't solve

The perimeter of an isosceles right triangle is 16+16√2 ((if you cant read that it is 16+16 * square root 2)). What is the length of the hypotenuse?

I understand that it should be 2x + x√2 = 16+16√2, but not sure how to solve it.

Any help would be great thanks!

Hello, fwl200!

The perimeter of an isosceles right triangle is: .
What is the length of the hypotenuse?

I understand that it should be: .
but not sure how to solve it.

The equal sides are ; the hypotenuse is


We have: .

Factor: .

Hence: .


Rationalize: .

. . . . . .


Therefore, the hypotenuse is: .

Let length of equal sides be =x
length of hypotenus be =y
according to the question
2x+y=16+16*2^(1/2)
also
y^2=2x^2 therfor
y=2^(1/2) x
substituting the value we get
2x+2^(1/2) x=16+16*2^(1/2)
therfor
x(2+2^(1/2))= 16+16*2^(1/2)
x=[16+16*2^(1/2)]/(2+2^(1/2))
rationalise it by multiplying numerator and denominator by (2-2^(1/2)) by doing so we get
x=8*2^(1/2)
since
y=x*2^(1/2)
so y=16
hence length of hypotenuse is 16 unit.

Thanks so much!