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An equilateral triangle ABC is inscribed in a circle as shown in the figure. A square of the largest possible area is made inside this triangle as shown. Another circle made inscribing the square. What is the ratio of the area of the small circle and the large circle?

An equilateral triangle ABC is inscribed in a circle as shown in the figure. A square of the largest possible area is made inside this triangle as shown. Another circle made inscribing the square. What is the ratio of the area of the small circle and the large circle? Correct Answer (63 - 36√3) ∶ 4

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Given:

As shown in the above figure the ΔABC is the given equilateral triangle and DECG is a square. Point O is the center of both circles.

Formula Used:

Area of circle = π × R2, where R is the radius of the given circle

Also the circumradius of a given equilateral triangle with side x unit = x/√3

Calculation:

Let the side of the equilateral triangle be A unit, and the radius of the bigger circle be R

Let the side of the square be a unit, then the radius of the inner circle will be a/2 units (As DECG is a square)

Now, in ΔDBG, tan60° = DG/BG

⇒ √3 = a/BG, BG = a/√3

Similarly the side FC = a/√3

Now the side of the equilateral triangle(A) will be given by the sum of BG + GF + FC

⇒ Side A = a/√3 + a/√3 + a

⇒ (√3a + 2a)/√3      ----(1)

Now by using the above-mentioned formula the value of the circumradius of the ΔABC is = A/√3

The radius of the bigger circle R

⇒ (√3a + 2a)/√3×√3 = (√3a + 2a)/3      ----(2)

Now by using the above-mentioned formula the area of the bigger circle is πR2 and of the smaller is π(a/2)2

The ratio of both the circles is πR2:π(a/2)2 

⇒ 4R2:a2 = 4{a(√3 +2)/3}2/a2 = 4(7 +4√3)/9 By rationalising

⇒ {(7 + 4√3)(7 – 4√3)×4}/9(7 – 4√3) = 4/(63 – 36√3)

∴ The ratio of the area’s of the two circles is 4/(63 – 36√3)

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