Area of a cyclic quadrilateral.

Question:

The distance $SR$ from $PQ$ is 7cm and arc $SR$ is 48cm and arc $SP \cong$ arc $QR$. Then find the area of quadrilateral $SRQP$($PQRS$ are taken in order and $O$ is centre).

What we(me and my friends) tried:

Approach 1:

In construction,$OM\perp PQ$.
Let radius of circle be $r$.
By Pythagoras theorem:
$$SM=MR=\sqrt{r^2-49}$$ $$SR=2\sqrt{r^2-49}$$ $$\text{Area} \triangle SOR =7\sqrt{r^2-49}$$ $$ \text{Area} \triangle SOP=\triangle ROQ=\dfrac{7r}{2}$$ Area of quadrilateral:
$$7r+7\sqrt{r^2-49}$$ Now $\angle OMR=\angle OMS=\theta$
$$\angle SOR=2\theta$$

By using radian arc formula: $$48=r\cdot 2\theta\dfrac{ \pi}{180}$$ $$\theta=\dfrac{4320}{\pi r}$$ In $\triangle OMR$:
$$\cos(\theta)=\dfrac{7}{r}$$ $$\cos\bigg(\dfrac{4320}{\pi r}\bigg)=\dfrac 7r$$ I have no idea how to simplify this.

Approach 2: Let $MN$ be $x$,$\angle SOR=\theta,\angle ROQ=\angle SOP=\phi$ and $\phi=\dfrac{180-\theta}{2}$
Radius of circle:
$$ON=OM+MN=7+x$$ $$\text{Area}\triangle SOR=\dfrac 12 (7+x)^2 \sin \theta$$ $$\text{Area}\triangle SOP=\text{Area}\triangle ROQ=\dfrac 12 (7+x)^2 \sin \phi$$ $$\text{Area of quadrilateral }PQRS=\dfrac 12 (7+x)^2 \sin \theta+ (7+x)^2 \sin \phi$$ $$=\dfrac 12 (7+x)^2 \sin \theta+ (7+x)^2 \sin \bigg(\dfrac{180-\theta}{2}\bigg)$$

And $$\frac \theta {360}[2\pi(7+x)]= 48$$ Two equations and two variables, so it might be solved( but I was not able to do so ).

How to solve this question?

Thanks!

As per comments, it should be solved by numerical approximation.

HINT.-Putting angle $\angle{SOR}=2\theta$ you have the system $$r\theta=24\\r\cos(\theta)=7$$ so you have the trascendental equation $7\theta=24\cos(\theta)$. An approximate solution is $\theta\approx 1.21$ so $r\approx\dfrac{24}{1.21}\approx19.83471$.

Now $SR=2r\sin(\theta)$ and the required area $A$ is given by $$A=\frac{7(PQ+SR)}{2}=\frac{7(2r+2r\sin(\theta))}{2}$$