# Find the area of the equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a another square of side 8 cm

find the area of the equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a another square of side 8 cm.

find the area of the equilateral triangle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a another square of side 8 cm.

Answer: Let’s call the square with side 8 cm “Square ABCD”, and let E, F, G be the midpoints of AB, BC, CD respectively, and H be the midpoint of AD. We can draw a line from E to G to divide Square ABCD into two equal parts. Let M be the point where the line from E to G intersects FH.

Now, we know that EM = MG = FG/2 = 4 cm (since F is the midpoint of BC). Also, EH = HG = GH/2 = 4 cm (since H is the midpoint of AD).

So, we have a rhombus EGHF with diagonals EG and FH both equal to 8 cm. The diagonals of a rhombus are perpendicular bisectors of each other, so EM and HG are perpendicular to each other at M. Therefore, the triangles EMH, EGM and GMF are all right-angled triangles.

Since the radius of the circle circumscribed by the square ABCD is equal to half the length of its diagonal (which is 8√2 cm), the radius of the circle is 4√2 cm.

Now, consider the right-angled triangle EGM. Since it is a 30-60-90 degree triangle with hypotenuse 4 cm, the length of the shorter leg EG is 2√3 cm.

Since the equilateral triangle inscribed in the circle has three equal sides, each side will be twice the length of EG, i.e. 4√3 cm. Therefore, the area of the equilateral triangle is:

(√3 / 4) x (side)^2
= (√3 / 4) x (4√3)^2
= (√3 / 4) x 48
= 12√3 sq cm

Thus, the area of the equilateral triangle inscribed in a circle circumscribed by a square made by joining the midpoints of the adjacent sides of another square of side 8 cm is 12√3 sq cm.
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