Maximum and minimum possible area of parallelogram

Question

$ABCD$ is a parallelogram with $AD=6$ , $CD=8$ , the angle between $AD$ and $CD$ is a variable

What is the Maximum and minimum possible area of parallelogram ?

My think about Maximum area is when the parallelogram turn to be a rectangle and so the Area = $4\times 8 = 48$

But what a bout the minimum possible area ?

Thanks for help !

Answer 1

Let's say $x=\angle BAD$. Then the height of the parallelogram to side $AD$ is $8\sin x$ and we have this formula for the area: $$A=6\cdot 8\cdot \sin x$$ Obviously, the maximum area is when $\sin x =1$, minimum is when $\sin x=0$. So the minimal area is zero.

Answer 2

The minimal possible area does not exist.

Indeed, let $S$ be a minimal possible area of the parallelogram and $\measuredangle ADC<90^{\circ}.$

Thus, $$S=6\cdot8\cdot\sin\measuredangle ADC,$$ which gives $$\measuredangle ADC=\arcsin\frac{S}{48}.$$

Now, let $0^{\circ}<\measuredangle ADC<\arcsin\frac{S}{48}.$

Thus, $$S=48\sin\measuredangle ADC<48\sin\arcsin\frac{S}{48}=S,$$ which is a contradiction.

Id est, our assumption that $S$ is a minimal possible area of the parallelogram was wrong.

Answer 3

It is possible to think of AD and CD as vectors centered at origin on the XY plane.

$\lvert \vec {DA}\rvert = 6 \\ \lvert \vec{DC}\rvert=8$

Let the angle between them be $\theta$.

The magnitude of the vector of the cross product of $\vec{DA}$ and $\vec{DC}$ will be equal to the area of the parallelogram.

$\lvert\vec{DA} \times \vec{DA}\rvert = \lvert \vec {DA}\rvert \lvert \vec {DC}\rvert \sin \theta$

$\lvert\vec{DA} \times \vec{DA}\rvert = 6 \cdot 8 \sin \theta$

$\lvert\vec{DA} \times \vec{DA}\rvert = 48 \sin \theta$

Since $\theta \in [0,\pi]$ and the area of the parallelogram's area depends on $\sin\theta$

The minimum value of the area of the parallelogram is $0$ and the maximum is $48$.