# 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 $$\phi=\dfrac{180-\theta}{2}$$
$$ON=OM+MN=7+x$$ $$\text{Area}\triangle SOR=\dfrac 12 (7+x)^2 \sin \theta$$ $$\text{Area}\triangle SOP=\text{Area}\triangle ROQ=\dfrac 12 (7+x)^2 \sin \phi$$ $$\text{Area of quadrilateral }PQRS=\dfrac 12 (7+x)^2 \sin \theta+ (7+x)^2 \sin \phi$$ $$=\dfrac 12 (7+x)^2 \sin \theta+ (7+x)^2 \sin \bigg(\dfrac{180-\theta}{2}\bigg)$$

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.

• You wrote $SR = 2\sqrt{r^2-49}$. You can equate that to $48$ and calculate $r$. With the radius known, finding the area of the quadrilateral will be easy. Nov 21, 2020 at 4:15
• The numbers are too perfect for that to be arc $SR$. However I will investigate that possibility, since it was written in your question. Nov 21, 2020 at 4:19
• Please check the original question. I strongly suspect that it should just be $SR = 48$, which has also been mentioned by player3236. Nov 21, 2020 at 4:21
• Whenever you have segment and arc involved, it is messy. If you know the angle and find arc or segment, it is fine but if arc and segments are known and you have to find subtended angle, it is not possible by hand without some guesses / approximation. Btw your equation in the first approach should just be $cos (\frac{24}{r}) = \frac{7}{r}$. Radius comes to $\approx 20 \,$ cm and angle subtended at the center by the arc is $\approx 140^0$. Nov 21, 2020 at 5:47
• Keeping the angle in radian. Arc of length $r$ will subtend an angle of $1$ radian ($2\pi r$ length subtends an angle of $2\pi$). Nov 21, 2020 at 11:56

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 $$A=\frac{7(PQ+SR)}{2}=\frac{7(2r+2r\sin(\theta))}{2}$$
• $1.21 radians=69.32789 degrees$ Dec 15, 2021 at 22:38