AMC-8 USA Math Olympiad

Geometry of circles and rectangles AMC 8 2014 problem 20

Try this beautiful problem from AMC 8. It involves geometry of circles and rectangles. We provide sequential hints so that you can try the problem.

What are we learning ?

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.

First look at the knowledge graph.

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]

Source of the problem
American Mathematical Contest 2014, AMC 8 Problem 20

Key Competency

Geometry of circles and rectangles 

Difficulty Level
Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

Start with hints 

Do you really need a hint? Try it first!
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.
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 $15-\frac{7\pi}{2}$.
Now what can we do with  $15-\frac{7\pi}{2}$ to get an approximate value ?
As we know that we can approximate \( \pi \) by \( \frac{22}{7} \) .  and substituting that in will give 15-11=4.

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