A geometry problem by Yahia El Haw

Since the perimeter is the sum of the side lengths, we have

$44=2(3x+2)+2(5x+4)$

$44=6x+4+10x+8$

$x=2$

Therefore, $3x+2=8$ and $5x+4=14$ .

The area of a parallelogram is $ab~\sin \theta$ where $a$ and $b$ are two adjacent sides and $\theta$ is the included angle. We have

$64=8(14)(\sin T)$

$\sin T=\dfrac{4}{7}$

$T=\sin^{-1} \approx \boxed{34.85^\circ}$

The parameter is given by:

$\begin{aligned} 2(3x+2 + 5x+4) & = 44 \\ 16 x + 12 & = 44 \\ \implies x & = 2\end{aligned}$

The area is given by:

$\begin{aligned} (3x+2)(5x+4)\sin(T) & = 64 \\ (6)(14)\sin(T) & = 64 \\ \sin(T) & = \frac 47 \\ \implies T & = \boxed{34.850}^\circ \end{aligned}$

From the perimeter, solve for $x$ , we have

$44=2(3x+2)+2(5x+4)$ $\implies$ $44=6x+4+10x+8$ $\implies$ $x=2$

The sides therefore area $3(2)+2=8$ and $5(2)+4=14$ . Now using the formula: $A=ab \sin \theta$ where $a$ and $b$ are side lengths and $\theta$ is the included angle, we have

$64=14(8) \sin~T$

$\sin~T=\dfrac{4}{7}$

$T=\sin^{-1}\dfrac{4}{7}=34.85^\circ$