# 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!

• 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. – player3236 Nov 21 '20 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. – player3236 Nov 21 '20 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. – Toby Mak Nov 21 '20 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 $\approx 140^0$. – Math Lover Nov 21 '20 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 $2\pi$). – Math Lover Nov 21 '20 at 11:56