An equilateral triangle ABC is inscribed in a circle as shown in figure. A square of largest possible area is made inside this triangle as shown. Another circle made inscribing the square. What is the ratio of area of large circle and the small circle?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is Option 2 i.e. (63 − 36√3) : 4
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 areas of both the circles:
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)