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## Sequential Hints:

Step 1:

Draw the DIAGRAM with necessary Information, please! This will convert the whole problem into a picture form which is much easier to deal with.

Step 2:

Power of a Point – Just the similarity of $$\triangle QOS$$ and $$\triangle POR$$

By the power of a point, PO . OQ = SO . OR . We know SO = 4; PO = 3.

Let, OQ be $$x$$. Hence we get the following:

SO = 4; PO = 3; OQ = $$x$$; OR = $$\frac{3x}{4}$$.

Step 3:

Assume $$\angle QOS = \theta$$ .

Now, compute the area in terms of $$x$$.

Area of $$\triangle QOS = 2x\sin{\theta}$$.

Area of $$\triangle POR = \frac{9x\sin{\theta}}{8}$$.

Therefore, we get the following that $$\frac{\triangle QOS }{\triangle POR } = \frac{16}{9}$$.

Hence the Area of $$\triangle QOS = \frac{112}{9}$$.