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Learn MoreTry this beautiful Problem on Geometry from Circular arc from (AMC 10 A, 2012).

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2 \pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2 .$ What is the area enclosed by the curve?

,

- $2 \pi+6$
- $2 \pi+4 \sqrt{3}$
- $3 \pi+4$
- $2 \pi+3 \sqrt{3}+2$
- $\pi+6 \sqrt{3}$

Geometry

Circle

Hexagon

Pre College Mathematics

AMC-10A, 2012

$\pi+6 \sqrt{3}$

We have to find out the area enclosed by the curve. but we can not find out eassily. Now we join the centre of 9 congruent circles. it will form a hexagon.

Can you find out the required area of the closed curve? Can you finish the problem?

The area of the Hexagon = $(6)(0.5)(2 \sqrt{3})=6 \sqrt{3}$. But we have to find out the area area enclosed by the curve. Then add the areas of the three sectors outside the hexaqon and subtract the areas of the three sectors inside the hexagon but outside the fiqure to get the area enclosed in the curved figure

The areas of the three sectors outside the hexaqon=$2 \pi$ .

the areas of the three sectors inside the hexagon but outside the figure $(\pi)$

Therefore the area enclosed by the curve is $\pi+6 \sqrt{3}$

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=k3eXbgwcNRw

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