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Find the attached sketch and explanation

The triangle that connects the centre of each smaller circle (r=2) is an equilateral triangle with side of 4.

The big circle that circumscribes the above figure has a radius of R. This radius R divides each corner angle of the equilateral triangle into 2 equal part (2*30 deg).

cos 30 deg = 2 /(R-2) = √3 /2
--> R = 4/√3 + 2

Area of big circle = π. R^2
= π.(4/√3 + 2)^2
= π/3.(4 + 2√3)^2

FINAL ANSWER IS (B)

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Image
Three circles with radius 2 cm are tangent to each other as shown above. What is the area of the circle, circumscribing the above figure?

A. \(\frac{π}{4}(4+2√3)^2\)

B. \(\frac{π}{3}(4+2√3)^2\)

C. \(3π(4+√3)^2\)

D. \(4π(4+√3)^2\)

E. \(5π(4+√3)^2\)
Refer diagram.
Attachment:
Three circles.png
Three circles.png [ 120.01 KiB | Viewed 8408 times ]
Sides of triangle in blue = 4 so it an equilateral triangle.
Here CD is perpendicular to side AB. CD = \(\sqrt{4^2 - 2^2}\) {from triangle CDB}
CD = \(2\sqrt{3}\)

Now, length of CO = \(\frac{2}{3}* any perpendicular = \frac{2}{3}*2\sqrt{3} = \frac{4}{\sqrt{3}}\) {where centroid lies \(\frac{2}{3}\) from any vertex, here C}

Radius of bigger circle = PC + CO = \(2 + \frac{4}{\sqrt{3}}\)
Area of bigger circle = π * r^2 = π * \((2 + \frac{4}{\sqrt{3}})^2\)
= \(\frac{π}{3}\) * \((2\sqrt{3} + 4)^2\)

Answer B.

Note: Since all the circles are of radius 2 and tangent to each other, the center of the figure of the three circles would coincide with centroid of the triangle formed by joining the centers of all the three circles.
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Quote:
Three circles with radius 2 cm are tangent to each other as shown above. What is the area of the circle, circumscribing the above figure?

A. \(\frac{π}{4}(4+2√3)2\)

B. \(\frac{π}{3}(4+2√3)2\)

C. \(3π(4+√3)2\)

D. \(4π(4+√3)2\)

E. \(5π(4+√3)2\)
Attachment:
img3.png
img3.png [ 18.53 KiB | Viewed 8211 times ]
if we join the center of the three circles we'll get a equilateral triangle.

we need a point that is equidistant from the three vertices. this point O is the circumcenter of the equilateral triangle.
OA = \(\frac{side}{\sqrt{3}}\) = \(\frac{4}{\sqrt{3}}\)

Radius of the bigger circle circumscribing the figure = OA + AP = \(\frac{4}{\sqrt{3}} + 2\) = \(\frac{4 + 2\sqrt{3}}{\sqrt{3}}\)
Area = \(\frac{\pi}{3}(4+ 2\sqrt{3})^2 \)
Ans: B
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Joining the centers of 3 circles, we get an equilateral triangle with each side 4 units

Circumradius of equilateral triangle = 2/3 * (height of triangle)

height of triangle =\( 2 * \sqrt{3}\)

Therefore, Circumradius = \(4/\sqrt{3}\)

Radius of outer circle = \(4/\sqrt{3}\) + 2 (circumradius + radius)


Area of outermost circle = Pi * (\(4/\sqrt{3}\) + 2) * (\(4/\sqrt{3}\) + 2)

On simplifying we get choice B


Answer choice: B
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The blue equilateral triangle can be divided into 6 congruent 30-60-90 triangles, as shown in the figure.
In each of the 30-60-90 triangles, the leg opposite the 60-degree angle is equal to the 2-cm radius of each smaller circle, with the result that the hypotenuse \(= \frac{4}{\sqrt{3}}\).
In the outer circle, the radius to point A \(= 2 + \frac{4}{\sqrt{3}} = \frac{2\sqrt{3} + 4}{\sqrt{3}}\).
Thus:
Area of the outer circle \(= π(\frac{2\sqrt{3} + 4}{\sqrt{3}})^2 = \frac{π}{3}(2\sqrt{3} + 4)^2\)

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Quote:
Three circles with radius 2 cm are tangent to each other as shown above. What is the area of the circle, circumscribing the above figure?

A. π4(4+2√3)2π4(4+2√3)2
B. π3(4+2√3)2π3(4+2√3)2
C. 3π(4+√3)23π(4+√3)2
D. 4π(4+√3)24π(4+√3)2
E. 5π(4+√3)2

find the circle that circumscribes all three smaller circles
find the distance from small circle origin to large circle origin
join the small origins, creating an equilateral triangle side 4
find the distance of perp bisector of the equi's side to large origin
this forms a 30-60-90 triangle with ratio's x:xV3:2x
the distance we want is the hypotenuse: 2x
60 is the radius of small circle = 2 = xV3, x=2/V3=2V3/3
90 is 2x=4V3/3
radius large circle is radius small circle + hypotenuse
radius large circle = 4V3/3 + 2 = 4V3+6/3
area large circle = (4V3+6/3)^2*pi

ans (B)
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Please See The Attached File.

We got an equilateral triangle by connecting the centers of three small circles.

The area = \((\frac{√3}{4})4^{2}= 3(\frac{4h}{2} )\)

\(4√3= 6h\)

--> \(h = \frac{2}{√3}\)

\(r = √((\frac{2}{√3})^{2}+ 4)= √\frac{16}{3}= \frac{4}{√3} \)

---> The center of outer circle of that equilateral triangle is the center of that big circle too.
And the radius of that big circle \(R = r +2\)

--->\(R= r +2= \frac{4}{√3}+ 2= \frac{(4+2√3)}{√3}\)

The area of Big Circle = \(πR^{2}= π\frac{(4+2√3)^{2}}{(√3)^{2}}= \frac{π}{3}(4+2√3)^{2}\)

Answer (B).
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