The side length of the regular triangle ABC is 12.
∴ BD = 1/2 BC = 6
∵ Circle O is tangent to AB and BC.
Bo equal division ∠ABC
∴ △BOD is Rt△, ∠ OBD = 30, ∠ BOD = 60.
∴ circle o radius od = BD× tan 30
= 6 × √3/3 = 2√3.
The circumference of the circle O is 2π × 2√3.
∵ The central angle of arc DG pair is 60.
The length of ∴ arc DG is 0/6 of the circumference of 65438+ circle O, which is 2√3 /3 π.
In order to find the arc length FG, we need to ask the radius r of the circle S.
∫SF = r∠SBF = 30
∴ BS = 2r
∴ BG = 3r
∫BG+OG = 20d OD = OG
∴BG+od = 20d
∴ Blood sugar = outer diameter
∴ 3r = OD
∴r = 1//3 od = 1/3×2√3 = 2√3/3
The circumference of the circle S is 2π × 2√3/3.
∵ The central angle of arc FG pair is 120.
The length of arc FG is 0/3 of the circumference of circle S, which is 4√3/9 π.
Find the area of the shaded part:
S △BSF = 1/2 × BF × SF
= 1/2 × ( √3SF ) × SF ( SF = 2√3 / 3)
= 2√3 / 3
The overlapping sector area of circle S and △BSF is 1/6 of the area of circle S,
Is it 1/6 × π × (2√3/3)? = 2/9 π
Shaded area: 2×(s△BSF- overlapping sector)
= 2 × ( 2√3 / 3 - 2/9 π )
= 4/9 ( 3√3 - π )