Here is a problem based on the area of triangle from ISI B.Stat Subjective Entrance Exam, 2018.
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 \(\triangle QOS = \frac{112}{9}\).
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