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

Question

Maverick Smith · Accepted Answer

Perimeter equals to #12sqrt(3)#

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)#.

Caroline Aucoin · Answer

Perimeter equals to #12sqrt(3)#

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)#

Wyatt Anderson · Answer

undefined The radius $ r $ of a circle inscribed in an equilateral triangle is related to the side length $ s $ of the equilateral triangle by the formula $ r = \frac{s\sqrt{3}}{6} $.

Given that $ r = 2 $, we can solve for the side length $ s $:
$$ 2 = \frac{s\sqrt{3}}{6} $$
$$ 12 = s\sqrt{3} $$
$$ s = \frac{12}{\sqrt{3}} = 4\sqrt{3} $$

The perimeter $ P $ of the equilateral triangle is three times the side length:
$$ P = 3s = 3(4\sqrt{3}) = 12\sqrt{3} $$

So, the perimeter of the equilateral triangle is $ 12\sqrt{3} $.