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2D Geometry – Areas related to circle AMC 8 2017 Problem 25

Try this beautiful problem from AMC 8. It involves calculating the area of a sector of a circle. We provide sequential hints so that you can try the problem.

Competency in Focus: 2D Geometry (Areas related to circle)

This problem from American Mathematics Contest 8 (AMC 8, 2017) is based on calculation of areas related to circle. It is Question no. 25 of the AMC 8 2017 Problem series.

Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 2, and $m\angle TUS = 60^\circ$. Arcs $TR$ and $SR$ are each one-sixth of a circle with radius 2. What is the area of the region shown? $\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][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=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”on”]American Mathematical Contest 2017, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]

Finding the area of a triangle and sector of a circle. (Area related to circles)

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]C0nstruction : Let $X$ and $Y$ are the centres of the scetors $ST$ and $TR$ Now Let us join $SX$ and $TY$ What do you think? Will the points $U,S,\textbf{ and}\quad X$ be in a straightline?[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]$U,S,\textbf{ and}\quad X$ will be in a straight line because $\angle STU =60^{\circ}$ And angle of a  circle is $360$  i.e., $\angle SXR = \angle TYR = 60^{\circ}$ [Since sector($SXR$)=$\frac{1}{6}circle$] Then $UXY$ will make an equilateral triangle.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]So after construction the figure will look like this : Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$.  [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$[/et_pb_tab][et_pb_tab title=”Formulas Used ” _builder_version=”4.2.2″]Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle][/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]