The radius of the circle is found to be 2 cm.
A triangle is said to be equilateral if all three of its sides are the same length and each of its three angles measures 60 degrees.
The given triangle is equilateral with a side length of 2cm. This triangle is on the top of a square BCDE. The circle circumscribes through the A, D, and E of the given shapes.
Using the given details, first, draw the diagram. Let H is the circumcenter of ADE. In the diagram, draw the altitude of AG passing through the circumcenter and square as GF.
From the diagram, let GH be x. The side of the triangle is 2 cm and the altitude cut it into two equal halves. Then, BG=EF=1 cm, HF=2-x, and HE=x+√3
[tex]\begin{aligned}\sin \theta &= \frac{AG}{AB}\\\sin 60^{\textdegree} &=\frac{AG}{2}\\\frac{\sqrt{3}}{2}&=\frac{AG}{2}\\AG&=\sqrt{3}\end{aligned}[/tex]
By Pythagoras' theorem, from ΔHEF,
[tex]\begin{aligned}HE^{2}&=HF^{2}+EF^{2}\\\left(x+\sqrt{3}\right)^2&=(2-x)^2+1^2\\x^2+3+2\sqrt{3}x&=4-4x+x^2+1\\(4+2\sqrt{3})x&=5-2\\x&=\frac{2}{4+2\sqrt{3}}\\&=\frac{2}{2(2+\sqrt{3})}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}\\&=\frac{2-\sqrt{3}}{4-3}\\&=2-\sqrt{3}\end{alinged}[/tex]
The radius of the circle (HE) is 2-√3+√3=2.
The answer is 2 cm. Therefore, option C is correct.
To know more about Pythagoras' theorem:
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