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Explore the Back-StoryTry 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

Try 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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I have A Big DOUBT:

In Problem:It should be 6 congruent circular arcs, instead of 9 congruent circular arcs,

The closed curve in the figure is made up of 6 congruent circular arcs each of length 2Ï€/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?

Pl Check If Problem needs Correction OR AM I Wrong?

I realised the problem is All-Right.

It is made of circular sectors of circle of radius of 1;

Three Outside sectors Of Hexagon: each of reflex angles (4/3)*Ï€;

Three Inside Sectors Of Hexagon each of obtuse angle (2/3)* Ï€;

Area of HeHexagon =6X (1/2)*2*âˆš3=6*âˆš3;

Three ouside sectors are equivalent to two circle of radiuus 1; To be added in area of hexagon

AND Three inside sectors are equivalent to one circle of radius 1; to be substracted in the area of hexagon;

Area of Figure provided= 6*âˆš3+Ï€* (1)^2=6*âˆš3+1

Area of Figure provided= 6*âˆš3+Ï€* (1)^2=6*âˆš3+1 is to be read as

6*âˆš3+Ï€

i think it will be 9 congruent

Yes,I realised it is nine Congruent ones!