Question 563793
i can solve this using trigonometry
i don't know how to solve it otherwise,
using trigonometry i would do the following:
the area of the larger circle is equal to 12*pi
since the area of a circle is equal to pi*r^2, this means that:
the radius of the circle = sqrt(12)
the radii of the big circle in the diagram are:
DB, DC, DA
the radii of the small circle in the diagram are:
DE, DF, DG
these radii are also the apothems of the triangle.
the apothems of the triangle intersect the sides of the triangle at a right angle.
this forms 8 right triangles.
they are:
DBF, DFC, DCG, DGA, DAE, DEB
the hypotenuse of each of these 8 triangles are the radii of the large circle.
this makes their length equal to sqrt(12)
the angle of each of these triangles with a vertex at the center of the circle is equal to 60 degrees.
using trigonometry, you can calculate the distance of the legs of these triangles that form the perimeter of the larger triangle which is an equilateral triangle and is labeled ABC.
using right triangle DBF:
the hypotenuse is DB with a length of sqrt(12)
the angle is BDF which is 60 degrees.
the sine of angle BDF is equal to opposite / hypotenuse which is equal to BF / BD
since BD is equal to sqrt(12), then we get:
sine (BDF) is equal to BF / sqrt(12)
multiply both sides of this equation by sqrt(12) to get:
BF = sqrt(12) * sine(BDF)
since angle BDF is 60 degrees, we get:
BF = sqrt(12) * sine (60) which becomes:
BF = 3
2 * 3 is equal to the length of one of the sides of the equilateral triangle.
3 * 2 * 3 is equal to the perimeter of the equilateral triangle.
that perimeter is equal to 18.
the diagram is shown below:
<img src = "http://theo.x10hosting.com/2012/jan281.jpg" alt = "$$$$" />