Because the sector = two radii+arc length If the radius is r and the central angle of the sector is n, then the perimeter of the sector is c = 2r+nπ r ÷ 180. Edit the sector area formula in this paragraph. In a circle with radius r, the sector area opposite the central angle of 360 is the circle area S = π r 2. So the area of a sector with a central angle of N: S = nπ r 2÷360, such as a circle with a radius of 1cm, then the circumference of a sector with a central angle of 135: c = 2r+n π r ÷180 = 2×/kloc-0. Sector area: s = n π R2 ÷ 360 =135× 3.14×1360 =1.1775 (square centimeter). 2lR where l is arc length and r is radius, edit the arc length formula of this sector l=(n/ 180)*pi*r, where l is arc length, n is the central angle of the sector, pi is π, and r is the triangle area formula of the sector radius. If the base of a triangle is known and its height is H, then S = ah/2, and the three sides of the triangle, A, B, C, and the semi-circumference P. Then S = √ [P (P-A) (P-B)] (Helen's formula) (p=(a+b+c)/2. If the radius of the inscribed circle is r, the triangle area =(a+b+c)r/2. Let the three sides of a triangle be A, B and C respectively, and the radius of the circumscribed circle be R, then the triangle area =abc/4r. Then s = √ {1/4 [C2A2-((C2+A2-B2)/2) 2]} (Qin Jiushao in Southern Song Dynasty) | A b 1 | S△ = 1. C d 1 | is a third-order determinant, and this triangle ABC is in the plane rectangular coordinate system A(a, b), B(c, d), C(e, f). It is better to choose ABC | e f 1 | from the upper right corner in counterclockwise order, because the results obtained in this way are generally positive. If you don't obey this rule, you might as well take it. The formula of circle area is: the radius of a circle is: r, the area is: s, then the area is S= π*r*r π, that is, the area of a circle is equal to pi times the radius of a circle times the radius of a circle, and the formula of bow area is: the arc opposite to bow AB is arc AB, then when arc AB is a bad arc, bow AB = s sector -s △ AOB (A, B are both ends of the arc. When arc AB is a semicircle, then S bow = S sector = 1/2S circle = 1/2× π r 2. When the arc AB is the optimal arc, the formulas of S-bow = S-sector+S △ AOB (A, B are the endpoints of the arc, and O is the center of the circle) are as follows: S = nπ R2 ÷ 360-ah ÷ 2S = π R2/2S = nπ R2 ÷ 360+.

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