##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 2.

##### Topic

Plane Geometry

##### Difficulty Level

5.5 out of 10

##### Suggested Book

‘Challenge and Thrill of Pre-College Mathematics’ by V,Krishnamurthy, C.R.Pranesachar, ect.

# Start with hints

Do you really need a hint? Try it first!

\(PQ\) and \( RS\) are two chords of the circle \(C\) , intersecting at the point \(O\). See figure: click here.

Given \(PO=3\) cm

\(SO=4\) cm

\([\triangle POR]= 7 cm^2\).

From the triangles \(POS\) and \(QOS\) we have,

\(\angle POR=\angle SOQ\) [Opposite angles]

\(\angle SRP=\angle SQO \) [Angle on the same semi-circle \(STP\)]

\(\angle QSO= \angle OPR\) [Angle on the same semi-circle \(ST’P\)]

Therefore the \(\triangle POR\) and \(\triangle SOQ\) are similar triangles .

\(\frac{[\triangle POR]}{OP^2}=\frac{[\triangle SOQ]}{SO^2}.\)

\(\Rightarrow [\triangle SOQ]=\frac{SO^2}{PO^2}\cdot [\triangle POR]\)=\(\frac{4^2}{3^2}\cdot 7=12\frac{4}{9}\).(Ans.)

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Area of △POR/Area of △QOS=(PO/SO)^2; putting known values;

or 7 cm 2/Area of △QOS=(3 cm/ 4 cm )^2=9/16;

or (7 cm ^2) *(16/9)=Area of △QOS;

Therefore Area of △QOS= (112/9) cm^2=12.4(recurring) cm ^2 answer