Geometry of circles and rectangles AMC 8 2014 problem 20

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What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.0.9" 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"]Competency in Focus: Geometry of circles and rectanglesThis problem from American Mathematics contest (AMC 8, 2014) will help us to learn more about geometry of circles and rectangles.

[/et_pb_text][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_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

First look at the knowledge graph.

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Next understand the problem

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Rectangle $ABCD$ has sides $CD=3$ and $DA=5$. A circle of radius $1$ is centered at $A$, a circle of radius $2$ is centered at $B$, and a circle of radius $3$ is centered at $C$. Which of the following is closest to the area of the region inside the rectangle but outside all three circles?[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw(Circle((0,0),1)); draw(Circle((0,3),2)); draw(Circle((5,3),3)); label("A",(0.2,0),W); label("B",(0.2,2.8),NW); label("C",(4.8,2.8),NE); label("D",(5,0),SE); label("5",(2.5,0),N); label("3",(5,1.5),E); [/asy][/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="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.0.9" open="on"]American Mathematical Contest 2014, AMC 8 Problem 20

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Geometry of circles and rectangles [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]6/10[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College MathematicsExcursion Of Mathematics 

[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.1" 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|0px|20px||" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints 

[/et_pb_text][et_pb_tabs _builder_version="4.1"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"]The area in the rectangle but outside the circles is the area of the rectangle minus the area of all three of the quarter circles in the rectangle.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"]Here the area of the rectangle is 3.5=15. Area of quater circles is (Area of the circle )/4 = \( \frac{\pi  . r^2}{4} \) , where r= radius of the circle . so, The area of all 3 quarter circles is $\frac{\pi}{4}+\frac{\pi(2)^2}{4}+\frac{\pi(3)^2}{4} = \frac{14\pi}{4} = \frac{7\pi}{2}$, where area of the quater for circle A is \( \frac{\pi}{4} \) ,for circle B is \( \frac {\pi .2^2}{4} \) , for circle C is \( \frac{\pi.3^2}{4} \).Therefore the area in the rectangle but outside the circles is $15-\frac{7\pi}{2}$.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]Now what can we do with $15-\frac{7\pi}{2}$ to get an approximate value ?[/et_pb_tab][et_pb_tab title="Hint 4 " _builder_version="4.1"]As we know that we can approximate \( \pi \) by \( \frac{22}{7} \) . and substituting that in will give 15-11=4.[/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"]

Connected Program at Cheenta

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Cheenta AMC Training Camp consists of live group and one on one classes, 24/7 doubt clearing and continuous problem solving streams.[/et_pb_blurb][et_pb_button button_url="https://cheenta.com/amc-8-american-mathematics-competition/" url_new_window="on" button_text="Learn More" button_alignment="center" _builder_version="4.0.9" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" background_layout="dark" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3"][/et_pb_button][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="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Similar Problems

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