1.

2.

3. Let's go through this step by step...can you find the area of $\displaystyle \triangle ABC$ ?

Originally Posted by MarkFL
Let's go through this step by step...can you find the area of $\displaystyle \triangle ABC$ ?
3200cm^2

5. Originally Posted by Yazan975
3200cm^2
That's not correct...how did you arrive at that answer?

Originally Posted by MarkFL
That's not correct...how did you arrive at that answer?
1600cm^2?

For 3200 I multiplied Base and Height but didn't divide by 2

7. Originally Posted by Yazan975
1600cm^2?

For 3200 I multiplied Base and Height but didn't divide by 2
Okay, good...so what must the area of $\displaystyle \triangle AED$ be?

Originally Posted by MarkFL
Okay, good...so what must the area of $\displaystyle \triangle AED$ be?
xy/2 cm^2

9. Originally Posted by Yazan975
xy/2 cm^2
Yes, that's correct in terms of $$x$$ and $$y$$, but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same.

This will give us an equation...can you state it?

Originally Posted by MarkFL
Yes, that's correct in terms of $$x$$ and $$y$$, but we should also be able to assign a numeric value to its area based on the fact that the areas of the two shaded regions are the same.

This will give us an equation...can you state it?
xy/2 cm^2 = 800
xy = 1600

Is that right?

11. Originally Posted by Yazan975
xy/2 cm^2 = 800
xy = 1600

Is that right?
Excellent!

Okay, now the next thing I notice is that within $\displaystyle \triangle ABC$ there are two similar triangles, with $\displaystyle \triangle AED$ being the smaller of the two. This means we may state:

$\displaystyle \frac{y}{x}=\frac{40}{x+10}$

Do you see where this comes from?

What I would do here is solve both equations we now have for $$y$$, and equate the two results to get an equation in $$x$$...can you state this equation?