Area of trapezium with triangles in it

Question

I've got a trapezium ABCD, AC and BD cross in point P. I have to find the Area of ABCD if $Ar(ABP)$ = $9cm^2$ and $Ar(CPD) = 1cm^2$ I am not sure how to write that, so I would be very grateful if someone could explain me. :)

Answer 1

Hint: let $\alpha = \widehat{APB}$ then dividing the two equations: $$\frac{1}{2}PA\;PB\;\sin \alpha = 9 \\ \frac{1}{2}PC\;PD\;\sin \alpha = 1$$ gives $\frac{PA}{PC}\;\frac{PB}{PD} = 9$. But $\frac{PA}{PC}=\frac{PB}{PD}$ so it follows that $\left(\frac{PA}{PC}\right)^2 = 9$ thus $\frac{PA}{PC} = \frac{PB}{PD} = 3$.

Then $S_{PDA} = \frac{1}{2}\; PD\;PA\;\sin (\pi - \alpha) = \frac{1}{2}\cdot\frac{1}{3}\;PB\;PA\;\sin \alpha = \frac{1}{3}\;S_{PAB} = \cdots$

Answer 2

The answer is $16$. Let $AB=a$, $CD=c$ and let $h_1$ and $h_2$ be the heights of the lower and upper triangle, resp. We have $ah_1=18$, $ch_2=2$ and $a/h_1=c/h_2$. From the first equation we conclude $h_1=18/a$. Express $c$ and $h_2$ by $a$ and you'll find $h_2=6/a$ and $c=a/3$. Finally $$\frac{a+c}{2}(h_1+h_2)=16.$$

Edit: if in general the areas of the triangles are $p$ and $q$ resp., the total area will be $$(\sqrt{p}+\sqrt{q})^2.$$