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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 rectangles This problem from American Mathematics contest (AMC 8, 2014) will help us to learn more about geometry of circles and rectangles.

Next understand the problem

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

[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.1″]

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 Mathematics Excursion Of Mathematics

[/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″]

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Categories

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. This problem from American Mathematics contest (AMC 8, 2014) is based on simple counting of semicircles.[/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″]

Next understand the problem

[/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″]A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5 miles per hour, how many hours will it take to cover the one-mile stretch?
Note: 1 mile = 5280feet [/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=”4.0.9″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.0.9″]American Mathematical Contest 2014, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.0.9″ open=”off”]Geometry of circles[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.0.9″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.0.9″ hover_enabled=”0″][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=”4.0.9″]How many lanes the highway consists of ? 2 right! given highway is 40 feet wide .Then width of each lane will be 40/2=20 feet wide .[/et_pb_tab][et_pb_tab title=”Hint 2″ _builder_version=”4.0.9″]Look at the diagram .See that the radius of each semicircle will be 20 feet on which Robert must be riding his bike .Again see that each semicrcle covers 40 feet of highway i.e. the diameter of the semicircle .[/et_pb_tab][et_pb_tab title=”Hint 3″ _builder_version=”4.0.9″]Calculate the number of semicircles over the whole mile . Number of semicircles=(length of the highway covered in total by Robert)/(length of highway covered by each semicircle)=5280/40 [since 1 mile=5280 feet] =132.[/et_pb_tab][et_pb_tab title=”Hint 4″ _builder_version=”4.0.9″ hover_enabled=”0″]Where the semicircles full circles ,their circumference would be 2.$\pi$ r =2.$\pi$.20=40 $\pi$ feet (since r=radius=20 feet). Therefore the circumference of semicircles is half that, or 20.$\pi$ feet. [/et_pb_tab][et_pb_tab title=”Hint 5″ _builder_version=”4.0.9″ hover_enabled=”0″]Therefore over the stretch of hghway, Robert rides a total of 132.20.$\pi$=2640. $\pi$ feet equivalent to $\frac{ \pi}{2}$  mile( since 1 mile=5280 feet) Given Robert rides at 5 miles per hour.So, time required by Robert =distance travelled/rate=($\frac{\pi}{2}$ miles)/(5 miles per hour)= $\frac{\pi}{10}$hours.   [/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″]