Question

Hi all
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Answer 1

area of the circle- area of the triangle def
through this you can find it.
hope it helps

Answer 2

In order to Find the Area of the Shaded Region : First We need to Find the Area of the Circle Inscribed in it. But, We Require Radius of the Circle to Find the Area of the Circle.

In order to Find the Radius of the Circle, We need to Find the Altitude(Median) of the Equilateral Triangle.

We can Notice from the Figure that :

✿  AE is an Altitude

✿  BE is Half of the Side of the Equilateral Triangle (Because E is the Midpoint of the Side)

✿  (AE - BE - AB) form a Right Angled Triangle

⇒ (AB)² = (BE)² + (AE)²

$\mathsf{Given : AB = 2\sqrt{3}\;and\;As\;BE\;is\;Half\;of\;the\;Side : BE = \sqrt{3}}$

$\mathsf{\implies (2\sqrt{3})^2 = (\sqrt{3})^2 + (AE)^2}$

$\mathsf{\implies 12 = 3 + (AE)^2}$

$\mathsf{\implies (AE)^2 = 9}$

$\mathsf{\implies AE = 3}$

We know that :

✿  In an Equilateral Triangle: Altitudes are same as Medians and pass through the Center of Circle inscribed in it, Because the Centroid and Orthocentre of an Equilateral Triangle is same as the Center of the Circle inscribed in it.

✿  A Centroid divides the Median in the Ratio of 2 : 1 and as the Centroid is same as Center of Circle in Equilateral Triangle, We can say that the Radius of the inscribed Circle will be one-third of the Altitude(Median)

$\mathsf{\implies Radius\;of\;the\;Inscribed\;Circle = \frac{3}{3} = 1\;cm}$

We know that : Area of Circle is given by : πr²

⇒ Area of the Inscribed Circle = π

Now, Let us find the Area of the Given Equilateral Triangle

$\mathsf{We\;know\;that,\;Area\;of\;an\;Equilateral\;Triangle\;is\;given\;by : \frac{\sqrt{3}}{4}(a)^2}$

$\mathsf{\implies Area\;of\;the\;Given\;Equilateral\;Triangle = \frac{\sqrt{3}}{4}(2\sqrt{3})^2}$

$\mathsf{\implies Area\;of\;the\;Given\;Equilateral\;Triangle = \frac{\sqrt{3}}{4}(12)}$

$\mathsf{\implies Area\;of\;the\;Given\;Equilateral\;Triangle = 3\sqrt{3}\;cm^2}$

We know that : The Four Triangles created by the Lines connecting Midpoints of the Sides of the Original Triangle are Congruent.

It means Area of the Triangle inscribed in the Circle has Area which is one-fourth the Original Triangle.

$\mathsf{\implies Area\;of\;the\;Inscribed\;Triangle = \frac{3\sqrt{3}}{4}\;cm^2}$

Area of the Shaded Portion = Area of the Circle - Area of the Inscribed Triangle in the Circle

$\mathsf{\implies Area\;of\;the\;Shaded\;Region = \pi - \frac{3\sqrt{3}}{4}}$

$\mathsf{\implies Area\;of\;the\;Shaded\;Region = (3.141 - 1.299)}$

$\mathsf{\implies Area\;of\;the\;Shaded\;Region = 1.842\;cm^2}$