In a parallelogram $ABCD$ the diagonals intersect at $O$. The distances from $O$ to sides $BC$ and $CD$ measure $2$ and $3$ respectively. Calculate the area of the parallelogram if $\angle ABC = 135^o$ (Answer:$24\sqrt2$)
My progess:
$S_{BCD}= \frac{BC.2}{2}=BC\\ S_{OCD} = \frac{CD.3}2=\frac{3CD}{2}\\ S_{BOC}=S_{COD} \implies BC = \frac{3CD}{2}\\ S_{ABCD} = 4BC = 6CD$
...ome detail is missing and I didn't see the given angle function
Hint: drop the altitude $AH$ from $A$ to $DC$. What is the length of $AH$? What are the angles in $\triangle DAH$?Can you calculate the area of parallelogram knowing the base and the height?