1. ## Algebra Problem

An isosceles right-angled triangle has one side length of (X+2)cm and the length of the hypotenuse is 12cm. Find the perimeter of the triangle.

I got up to (x+2)^2 + (X+2)^2 = 144

And i also worked out a method with the right answer, 2(x+2)^2 = 144

could someone explain to me why the first one doesnt work?

2. If it is an isosceles right triangle, then both of the of the other angles are the same and both of the adjacent sides are the same length.

Using Pythagorean theory:

$a^2 = b^2 + c^2
$

Gives us:

$
12^2 = (x+2)^2 + (x+2)^2$

Which expands to:

$
2x^2 + 8x + 8 = 144$

Factor to get:

$x^2 + 4x + 4 = 72$

Subtract 72 from both sides:

$x^2 + 4x - 68 = 0$

Use the quadratic equation to solve for $x$

Remember that the answer has to be $>0$, so discard any negative solutions.

3. Originally Posted by andrew2322

An isosceles right-angled triangle has one side length of (X+2)cm and the length of the hypotenuse is 12cm. Find the perimeter of the triangle.

I got up to (x+2)^2 + (X+2)^2 = 144 Mr F says: Yeah, and that's the same as 2(x + 2)^2 = 144, which is what you have below ......

And i also worked out a method with the right answer, 2(x+2)^2 = 144 Mr F says: This will let you solve for x. But x is NOT the perimeter.

could someone explain to me why the first one doesnt work?
The hypotenuse is 12. It's isosceles so the other two sides HAVE to be (x + 2). You want the perimeter: P = 2(x + 2) + 12 = 2x + 16.

So all you need now is the value of x. You get that from what you've already done ..... $x = \sqrt{72} - 2 = 6\sqrt{2} - 2$. So P = ......

4. Originally Posted by topher0805
If it is an isosceles right triangle, then both of the of the other angles are the same and both of the adjacent sides are the same length.

Using Pythagorean theory:

$a^2 = b^2 + c^2
$

Gives us:

$
12^2 = (x+2)^2 + (x+2)^2$

Which expands to:

$
2x^2 + 8x + 8 = 144$

Factor to get:

$x^2 + 4x + 4 = 72$

Subtract 72 from both sides:

$x^2 + 4x - 68 = 0$

Use the quadratic equation to solve for $x$

Remember that the answer has to be $>0$, so discard any negative solutions.
No need to expand and use quadratic formula:

$2(x + 2)^2 = 144 \Rightarrow (x + 2)^2 = 72 \Rightarrow x + 2 = \sqrt{72} \Rightarrow x = \sqrt{72} - 2 = 6\sqrt{2} - 2$.

Note: -ve root solution was discarded before even appearing since x > 0.

5. Originally Posted by mr fantastic
No need to expand and use quadratic formula:

$2(x + 2)^2 = 144 \Rightarrow (x + 2)^2 = 72 \Rightarrow x + 2 = \sqrt{72} \Rightarrow x = \sqrt{72} - 2 = 6\sqrt{2} - 2$.

Note: -ve root solution was discarded before even appearing since x > 0.
Yeah you're right I didn't notice that. Thanks for the heads up.