# Thread: Integrated Geometry and Algebra

1. ## Integrated Geometry and Algebra

Hi all,

In diag belo I have to find length of red triangle.

1). Equation x^2 + y^ = 9 Do i make y^2 subject here? And place in y^2 below:

(12- x^2)^2 - (9- y^2)^2 = c^2 ?

2. ## Re: Integrated Geometry and Algebra

?? The red portion is a rectangle, not a triangle.

You started out the way I would but you have only two equations in three unknowns, x, y, and c. You need a third, independent, equation.

3. ## Re: Integrated Geometry and Algebra

Yes, sorry for the error. It's length of red rectangle.. How do I obtain 3rd equation?

4. ## Re: Integrated Geometry and Algebra

The three equations:
$x^2+y^2=3^2$
$(12-x)^2+(9-y)^2=z^2$
$xy+(12-x)(9-y)+3z = 12\cdot 9$

The last equation comes from adding up areas.

This gives $z \approx. 12.2553$

5. ## Re: Integrated Geometry and Algebra

You can also use similar triangles:
$\dfrac{y}{12-x} = \dfrac{x}{9-y} = \dfrac{3}{z}$