Here is a problem based on the area of triangle from ISI B.Stat Subjective Entrance Exam, 2018.

## 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 \(\triangle QOS = \frac{112}{9}\).