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!
