In the diagram to the right, equilateral triangle ADE is

Refer diagram below

If I get you correct, you are talking about the red lines drawn bisecting the 60 deg angle to 30 deg

Yes ; that's correct because the CG of inscribed circle = CG of equilateral triangle

The red line drawn is joining the midpoint of the opposite side; so it will bisect; That's the rule

Yes, exactly! So because the center of the circle = center of triangle the line bisects the angle?

For this problem, we can summarize as follows

1. The inscribe circle touches the midpoints of all sides of the equilateral triangle
2. The lines in red are the medians of the equilateral triangle
3. Medians also bisect the angle 60 of equilateral triangle to angle 30
4. In an equilateral triangle, the median, angle bisector, and altitude are equal. They intersect at center of the inscribed and circumscribed circle.

Agree.. This is correct

Say r=1 and O the center of the circle. Then because of ADE=60, OD biscepts ADE to 30+30.
Therefore we have a triangle ODF with OF=1 and FD= √3 . (OF is the line that biscepts ED to its half).
We know that 30-60-90 triangle have sides 1-√3 - 2.

Ac=πr^2=π*1=π
Asq=ΕD^2= (2 √3 ) ^2 = 12

Ac/Asq=π/12

Let the side of the square is X, and AP is the perpendicular distance from vertex A to side ED.
Side of the equilateral triangle is also X according to the figure
Half of an equilateral triangle(AEP) is 30-60-90 triangle, and sides are in the ratio of a:a√3:2a
a=X/2(half of the side of the equilateral triangle)
AP=√3X/2
Since it is inscribed circle, centriod and the center of the circle will be same. AP will be median, and centroid will divide it in 2:1(2Y:1Y)
3Y=AP=√3X/2
Y=X/2√3(Radius of the circle)
area of the circle/area of the square=pi*Y^2/X^2
=(pi*X^2/12)/X^2
=pi/12
A:)

Rule 1: the In-radius of an Inscribed Circle inside an Equilateral Triangle = (1/3) * (Height of Equilateral Triangle)

Since the question asks for a Ratio answer, we can choose Numbers to fit the Diagram.

Let Side of Equilateral Triangle = 1

The Side of the Equilateral Triangle on the Bottom shares the Side of the Square ------> thus Side of Square = 1

Area of Square = (1)^2 = 1


In-radius of Inscribed Circle:

Height of the Equilateral Triangle is given by = (side) * sqrt(3) * (1/2)

(1/3) * (Height of Equilateral Triangle) = Radius of Inscribed Circle

given side =1:

(1/3) * (1) * sqrt(3) * (1/2) = sqrt(3)/6 = Radius of Inscribed Circle

Area of Inscribed Circle = (pi) * (r)^2 = (sqrt(3) / 6)^2 * (pi) = 3(pi) / 36 = (pi)/12


Ratio of the: (Area of Circle) / (Area of Square) = [ (pi) / 12 ] / [1] = (pi) /12

-A-

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