How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# 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://www.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"]
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017
[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Plane Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]6 out of 10

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Do you really need a hint? Try it first!

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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 $\Rightarrow DC=6×2=12$.

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Now  DC is diameter of the circle $\Rightarrow \angle DEC=90^\circ$. Therefore $\angle DEA$ is also right angle.

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The length of $DE=\sqrt{5^2-4^2}=3$.  And the length of $EC=\sqrt{12^2-3^2}=3\sqrt{15}$. Therefore $AC=AE+EC=4+3\sqrt{15}$.

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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.)

## Connected Program at Cheenta

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.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problems

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# Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy