Four Points on a Circle, ISI Entrance 2017, Subjective Problem no 2
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[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.4"]Given, \(AD=5, AE=4\) and \(\angle DBC=90^\circ \).As \(D,B,C\) are points on the circle having radius 6.Therefore \(DC\) is the diameter of the circle\(\Rightarrow DC=6×2=12\).
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.4"]Now DC is diameter of the circle \(\Rightarrow \angle DEC=90^\circ \).Therefore \(\angle DEA\) is also right angle.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.4"]The length of \(DE=\sqrt{5^2-4^2}=3\). And the length of \(EC=\sqrt{12^2-3^2}=3\sqrt{15}\).Therefore \(AC=AE+EC=4+3\sqrt{15}\).
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.4"]From \(∆AED\) and \(∆ABC\) we have,\(\frac{AD}{DE}=\frac{AC}{BC} \Rightarrow BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}\).Therefore the length of \(BC\) is \(\frac{12+9\sqrt{15}}{5}\). (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" custom_margin="59px||48px|||" 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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