Areas of Circles
There are 3 congruent circles inscribed into an equilateral triangle. If one of the sides of the triangle is 6+2 long, what’s the sum of the areas of the 3 circles? Express in terms of pi.
-- James Yu
-- James Yu
Labels: mathschallenge, SharedPuzzle
posted by Rajesh Lal at
2:38 PM
12 Comments:
sum of the circles is 47.25
9pi
is it supposed to be 6+2, or 6+(2*sqrt 3?
9 pi
sum of the circles = 28.2744 = 9 pi
each circle has a radius of 1.7321
and an area of 9.4248
area of 3 circles = 9. Pi.
Suppose that radius of each circle is x
then we have equation : 2x+ 2*sqrt(3)=6+2.Sqrt(3)
=> x= sqrt(3)
=> area = 3* (3* sqrt(3)^2)=9* Pi.
Result : 9.Pi
bill_rain@yahoo.com
The above answer seems to be incorrect. It states:
2x + 2*sqrt(3) = (lenght of triangle side)
The terms in the equation:
2x = diameter
2*sqrt(3) = ?
It seems to me that radius was labled as x and then just assumed to be equal to 3 and then plugged into a formula that represents nothing.
Agreed! Consider what he wrote:
2x+ 2*sqrt(3)=6+2.Sqrt(3)
=> x= sqrt(3)
=> area = 3* (3* sqrt(3)^2)=9* Pi.
In the first line x is clearly = to 3.
In the second line x now equals the squareroot of 3.
And I am also curious how one comes to the conclusion that "diameter + 2*sqrt(3) = (tri side lenght)
The sum of the circles is (4 + (2/3)sqrt(3))pi
Each circle is tangent to the triangle at a distance of 1/3 its length. If you cut the triangle in half one circle will be tangent to new triangle. The distance from the 90 deg cut to the original tangent will be 1/6 the original length of the original triangle. This will be the circles radius. So
(6+ 2sqrt(3))/6 = radius
simplify to (3+sqrt(3))/3
r^2 = (9 + 2sqrt(3) + 3)/9 -> (12+ 2(sqrt(3))/9 for the area of all three circles we multiply it by 3pi and get ((12+ 2(sqrt(3))/3)pi or (4 + (2/3)sqrt(3))pi
it's 47.25
9Pi
the answer is 9 pi
this is the easiest question ever give me a real challenge
by the way the answer is 9 pi
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