Question

The radius of a circle inscribed in an equilateral triangle is 2. What is the perimeter of the triangle?

Answer 1

Perimeter equals to `12sqrt(3)`

Explanation:

There are many ways to address this problem.
Here is one of them.

The center of a circle inscribed in to a triangle lies on intersection of its angles' bisectors. For equilateral triangle this is the same point where its altitudes and medians intersect as well.

Any median is divided by a point of intersection with other medians in proportion `1:2`. Therefore, the median, altitude and angle bisectors of an equilateral triangle in question equals to
`2+2+2 = 6`

Now we can use Pythagorean theorem to find a side of this triangle if we know its altitude/median/angle bisector.
If a side is `x`, from Pythagorean theorem
`x^2 - (x/2)^2 = 6^2`

From this:
`3x^2 = 144`
`sqrt(3)x=12`
`x = 12/sqrt(3) = 4sqrt(3)`

Perimeter equals to three such sides:
`3x = 12sqrt(3)`.

Answer 2

Perimeter equals to `12sqrt(3)`

Explanation:

Alternative method is below.

Assume, our equilateral triangle is `Delta ABC` and it center of an inscribed circle is `O`.

Draw a median/altitude.angle bisector from vertex `A` through point `O` until it intersects side `BC` at point `M`. Obviously, `OM=2`.

Consider triangle `Delta OBM`.
It's right since `OM_|_BM`.
Angle `/_OBM=30^o` since `BO` is an angle bisector of `/_ABC`.
Side `BM` is half of side `BC` since `AM` is a median.

Now we can find `OB` as a hypotenuse in a right triangle with one acute angle equal to `30^o` and cathetus opposite to it equal to `2`. This hypotenuse is twice as long as this cathetus, that is `4`.

Having hypotenuse `OB` and cathetus `OM`, find another cathetus `BM` by Pythagorean Theorem:
`BM^2 = OB^2 - OM^2 = 16-4=12`

Therefore,
`BM=sqrt(12)=2sqrt(3)`
`BC = 2*BM = 4sqrt(3)`
Perimeter is
`3*BC = 12sqrt(3)`