Plane Geometry [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]
5.5 out of 10
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'Challenge and Thrill of Pre-College Mathematics' by V,Krishnamurthy, C.R.Pranesachar, ect.
Do you really need a hint? Try it first!
[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]\(PQ\) and \( RS\) are two chords of the circle \(C\) , intersecting at the point \(O\). See figure: click here.
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]Given \(PO=3\) cm\(SO=4\) cm\([\triangle POR]= 7 cm^2\).
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]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 .
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]\(\frac{[\triangle POR]}{OP^2}=\frac{[\triangle SOQ]}{SO^2}.\)\(\Rightarrow [\triangle SOQ]=\frac{SO^2}{PO^2}\cdot [\triangle POR]\)=\(\frac{4^2}{3^2}\cdot 7=12\frac{4}{9}\).(Ans.)
[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.
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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