Find the area of the smaller part of the circle into which it was divided

| May 1, 2021 | education

Find the area of the smaller part of the circle into which it was divided by the smaller diagonal of the hexagon inscribed in the circle, if this diagonal is 6√3 cm.

Let’s build the radii ОА, ОВ and ОВ.

Triangles AOB and BOC are equilateral, then the central angle AOC = 120.

The AOS triangle is isosceles, since OA = OC = R. OH is its height, bisector and median. Then AH = AC / 2 = 6 * √3 / 2 = 3 * √3 cm.

In a right-angled triangle AOH Sin60 = AH / OA.

ОА = АН / Sin60 = 3 * √3 / (√3 / 2) = 6 cm.

R = ОА = 6 cm.

Determine the area of ​​the circle. Scr = π * R ^ 2 = 36 * π cm2.

The area of ​​the OABS sector is the third part of the circle. Ssec = Scr / 3 = 36 * π / 3 = 12 * π cm2.

Determine the area of ​​the triangle AOC.

Saos = ОА * ОС * Sin120 / 2 = 6 * 6 * √3 / 4 = 9 * √3 cm2.

Then the area of ​​the ABC segment is equal to: Sseg = Ssec – Saos = 12 * π – 9 * √3 = 3 * (4 * π – 3 * √3) cm2.

Answer: The area of ​​a part of a circle is 3 * (4 * π – 3 * √3) cm2.



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