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AMC 8 Math Olympiad USA Math Olympiad

Area of the figure | AMC-8, 2014 | Problem 20

Try this beautiful problem from Geometry:Area inside the rectangle but outside all three circles.AMC-8, 2014. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on the Area of the figure.

Area of the figure – AMC-8, 2014- Problem 20


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?

Area of the figure
  • $3.5$
  • $ 4.0$
  • $4.5$

Key Concepts


Geometry

Rectangle

Circle

Check the Answer


Answer: $4.0$

AMC-8 (2014) Problem 20

Pre College Mathematics

Try with Hints


Area of the red shaded portion

To Find out the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure), we have to find out the area of the Rectangle -the area of three quarter circle inside the circle(i.e green shaded region)

Can you now finish the problem ……….

To find out the area of the rectangle, AD=5 and CD=3 are given. to find out the area of the three-quarter circles, the radii are 1,2 & 3 respectively.

Now area of Rectangle=\(AD \times CD\) and area of  quarter circles =\(\frac{\pi r^2}{4}\),where \(r\)=Radius of the circle

can you finish the problem……..

Area of the red shaded portion

Area of the rectangle=\( 5 \times 3\)=15 sq.unit

Area of the quarter circle with the center C=\(\frac{\pi (3)^2}{4}\)=\(\frac{9 \pi}{4}\) sq.unit

Area of the quarter circle with the center B= \(\frac{\pi (2)^2}{4}\) =\(\frac{4 \pi}{4}\)=\(\pi\) sq.unit

Area of the quarter circle with the center C= \(\frac{\pi (1)^2}{4}\) =\(\frac{\pi}{4}\) sq.unit

Therefore the area of the region inside the rectangle but outside all three circles,( i.e the red shaded region in the above figure) =(15- \(\frac{9 \pi}{4}\)- \(\pi\) – \(\frac{\pi}{4}\) )=15-\(\frac{7\pi}{2}\)=15-11=4 sq.unit (Taking \(\pi =\frac{22}{7}\))

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