Area of triangle inside parallelogram

[cjwalle]

Homework Statement

http://folk.uio.no/robinbj/gg/ggstart.pdf"
I am supposed to find the area of the triangle PQD. The numbers given are the areas of the other triangles.

Homework Equations

$$A= \frac{1}{2} a b \sin{C}$$

As well as Heron's formula, possibly?
$$A= \sqrt{s(s-a)(s-b)(s-c)}$$ where $$s = \frac{a+b+c}{2}$$

The Attempt at a Solution

Where I'm stumped is exactly how to start. At first, I figured I'd try to find the values of a, b and c. Using the area of a triangle:

$$a = DP = \frac{14}{AP\sin{\theta}}$$

Similarly, for b:

$$b = QP = \frac{56}{QB\sin{\alpha}}$$

And c:

$$c = QD = \frac{28}{QC\sin{x}}$$

Meaning that $$s = \frac{7}{AP\sin{\theta}} + \frac{28}{QB\sin{\alpha}} + \frac{14}{QC\sin{x}}$$

However, dealing with three different angles, without knowing the relationship between them or the sum of the angles, I just don't know how to proceed and solve the exercise based on this. I am not looking for a solution from you guys, mind you. Just a tip to get me on the right track?

Thank you.

 

[tiny-tim]

Hi cjwalle!

I don't think the angles matter …

won't the ratios be the same, whatever the angles are, so you might as well make it a square?