Thread: Geometry question I can't solve

1. 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!

2. Hello, fwl200!

The perimeter of an isosceles right triangle is: . $16+16\sqrt{2}$
What is the length of the hypotenuse?

I understand that it should be: . $2x + x\sqrt{2} \:=\: 16+16\sqrt{2}$
but not sure how to solve it.

The equal sides are $x$; the hypotenuse is $x\sqrt{2}$

We have: . $2x + x\sqrt{2} \;=\;16 + 16\sqrt{2}$

Factor: . $x(2 + \sqrt{2}) \;=\;16(1 + \sqrt{2})$

Hence: . $x \;=\;\frac{16(1+\sqrt{2})}{2 + \sqrt{2}}$

Rationalize: . $x \;=\;\frac{16(1+\sqrt{2})}{2+\sqrt{2}}\cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} \;=\;\frac{16(2 - \sqrt{2} + 2\sqrt{2} - 2)}{4 - 2}$

. . . . . . $x \;=\;\frac{16\sqrt{2}}{2} \;=\;8\sqrt{2}$

Therefore, the hypotenuse is: . $x\sqrt{2} \;=\;(8\sqrt{2})(\sqrt{2}) \;=\;\boxed{16}$

3. 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.

4. Thanks!

Thanks so much!