Parallelogram with extended side lengths

First, we will show that the area of any convex quadrilateral is $\frac{1}{2}\cdot x \cdot y\cdot \sin \theta$ , where $x$ and $y$ are the lengths of the diagonal and $\theta$ is the angle between them.

Let $PQRS$ be a convex quadrilateral. Let $PR$ and $QS$ intersect at $T$ . Let $[PQRS]$ denote the area of $PQRS$ . Using the fact $\sin \angle PTQ = \sin \angle QTR = \sin \angle RTS = \sin \angle STP$ , we have

$\begin{aligned} [PQRS] &= [PQT] + [QRT] + [RST] + [SPT] \\ &= \frac{1}{2} \cdot PT \cdot TQ \cdot \sin \angle PTQ + \frac{1}{2} \cdot QT \cdot TR \cdot \sin \angle QTR \\ &\ + \frac{1}{2} \cdot RT \cdot TS \cdot \sin \angle RTS + \frac{1}{2} \cdot ST \cdot TP \cdot \sin \angle STP \\ &= \frac{1}{2}\cdot QT \cdot (PT + TR)\cdot \sin \angle PTQ + \frac{1}{2} TS \cdot (RT + TP)\cdot \sin \angle PTQ \\ &= \frac{1}{2}\cdot (PT + TR)(QT + TS)\cdot \sin \angle PTQ \\ &= \frac{1}{2}\cdot PR \cdot QS\cdot \sin \angle PTQ \\ \end{aligned}$

So, the area of parallelogram $ABCD$ is $\frac{1}{2} \cdot AC \cdot BD \cdot \sin \theta$ , where $\theta$ is the angle between the diagonals $AC$ and $BD$ .

Note that $\theta$ is also the angle between the diagonals $AC'$ and $BD'$ . So, the area of the quadrilateral $ABC'D'$ is $\frac{1}{2} \cdot AC' \cdot BD'\cdot \sin \theta = \frac{1}{2} \cdot (1.2 AC) \cdot (0.9 BD) \cdot \sin \theta = \frac{1.08}{2} \cdot AC \cdot BD\cdot \sin \theta$ .

Thus the ratio of the areas is $\frac{1.08 \cdot AC \cdot BD \sin \theta }{AC \cdot BD \sin \theta} = 1.08 = \frac{27}{25}$ . Hence $a + b = 27 + 25 = 52$ .