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). Question figure

What we(me and my friends) tried:

Approach 1: My work

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: A-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?


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

  • 1
    $\begingroup$ 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. $\endgroup$
    – player3236
    Nov 21, 2020 at 4:15
  • 1
    $\begingroup$ The numbers are too perfect for that to be arc $SR$. However I will investigate that possibility, since it was written in your question. $\endgroup$
    – player3236
    Nov 21, 2020 at 4:19
  • $\begingroup$ Please check the original question. I strongly suspect that it should just be $SR = 48$, which has also been mentioned by player3236. $\endgroup$
    – Toby Mak
    Nov 21, 2020 at 4:21
  • 2
    $\begingroup$ 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 $\approx 140^0$. $\endgroup$
    – Math Lover
    Nov 21, 2020 at 5:47
  • 1
    $\begingroup$ Keeping the angle in radian. Arc of length $r$ will subtend an angle of $1$ radian ($2\pi r$ length subtends an angle of $2\pi$). $\endgroup$
    – Math Lover
    Nov 21, 2020 at 11:56

1 Answer 1


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 $$A=\frac{7(PQ+SR)}{2}=\frac{7(2r+2r\sin(\theta))}{2}$$

  • $\begingroup$ $1.21 radians=69.32789 degrees$ $\endgroup$
    – Piquito
    Dec 15, 2021 at 22:38
  • $\begingroup$ Thanks for answering :-) I was going to self answer it to close it. $\endgroup$
    – Wolgwang
    Dec 16, 2021 at 1:14
  • $\begingroup$ @Wolgwang: You are welcome. The "difficulty" in your problem is the trascendental equation you need to solve, nothing more. $\endgroup$
    – Piquito
    Dec 16, 2021 at 12:50

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