# Area of the triangle math

1. Jan 10, 2005

### footprints

The shorter sides of a right-angled triangle are of length (x+y)cm, (x-y)cm respectively. Given that the length of the hypotenuse is $$\sqrt68$$cm and that the area of the triangle is $$8cm^2$$, find x and y.

$$\displaystyle{(x+y)^2 + (x-y)^2 = (\sqrt68)^2}$$
$$\displaystyle{x^2+y^2=34}$$ ----------- (1)
$$\displaystyle{\frac{1}{2}(x+y)(x-y)=8}$$
$$\displaystyle{x^2+y^2=16}$$ ----------- (2)
I can't seem to solve it. Did I do something wrong?

2. Jan 10, 2005

### Galileo

Equation (2) should be $x^2-y^2=16$.

Can you solve it now?

3. Jan 10, 2005

### footprints

Why?
This is how I did it.
$$\displaystyle{\frac{1}{2}(x+y)(x-y)=8}$$
$$x^2-xy+xy+y^2=16$$
$$\displaystyle{x^2+y^2=16}$$
Does that mean that -y multiply by y equal to -y^2? Isn't it y^2?

4. Jan 10, 2005

$$-y \times y = -y^2$$ is correct. Maybe you're thinking of $$-y \times -y = y^2$$

5. Jan 10, 2005

### footprints

I was thinking of the first one. I always thought that because there is a square, so it would be positive.

6. Jan 10, 2005

### Zaimeen

No, it's negative!

7. Jan 10, 2005

### hypermorphism

-y*y = -1*y*y = -1*(y*y) = -1*$$y^2$$ = -$$y^2$$