Four Points on a Circle, ISI Entrance 2017, Subjective Problem no 2
Understand the problem
[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"] Consider a circle of radius 6 as given in the diagram below. Let \(B,C,D\) and \(E\) be points on the circle such that \(BD\) and \(CE\), when extended, intersect at \(A\). If \(AD\) and \(AE\) have length 5 and 4 respectively, and \(DBC\) is a right angle, then show that the length of \(BC\) is \(\frac{12+\sqrt{15}}{5}\). Diagram : click here. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" link_option_url="https://cheenta.com/isicmientrance/" link_option_url_new_window="on"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="4.3.4" hover_enabled="0"]'An excursion in Mathematics' published by Bhaskaracharya Pratishthana [/et_pb_accordion_item][/et_pb_accordion][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="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" inline_fonts="Aclonica"]
Start with hints
[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.7" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!
[/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"]Watch the video ( Coming Soon ... )
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