1, triangle

Area: s = ah/2

(2) Given three sides A, B and C of a triangle, then (Helen formula) (p=(a+b+c)/2)

S=√[p(p-a)(p-b)(p-c)]

=( 1/4)√[(a+b+c)(a+b-c)(a+c-b)(b+c-a)]

(3) Given the included angle C between a and b on both sides of a triangle, S = 1/2 * ABS Inc.

(4) Let the three sides of the triangle be A, B and C respectively, and the radius of the inscribed circle be R..

S=(a+b+c)r/2

(5) Let the three sides of the triangle be A, B and C respectively, and the radius of the circumscribed circle be R..

S=abc/4R

(6). Find the area according to the trigonometric function:

s = absinC/2 a/sinA = b/sinB = c/sinC = 2R

Note: where r is the radius of the circumscribed circle. Circumference: l=a+b+c2, circle

Area: s = π * r 2

=π*D^2/4

= l 2/4 π (d: diameter, l: circumference)

Circumference: l=2πR

=πD3, sector

Area: s = n π * r 2/360

= ar 2 (n: the central angle of the sector, a: the arc system of the central angle of the sector)

Circumference: l=nπR/ 180+2R

=aR+2R4, ellipse

Area: S=abπ5, squared

Area: s = a 2

Circumference: l=4a6, rectangular

Area: S=ab

Circumference: l=2(a+b)7, parallelogram

Area: S=ah

=absinx (a: bottom, h: height, b: adjacent to a, x: angle between a and b)

Circumference: l=2(a+b)8, rhombic.

There are other formulas that apply to parallelograms:

Area: S=ab (a and B are the lengths of two diagonal lines)

Circumference: l=4x (x is the side length) 9. trapeziform

Area: S=(a+b)h/2 (a, b is the top and bottom, and h is the height)

Area of isosceles trapezoid: S=csinA(a+b)/2 (c is waist and a is acute angle) 10, circle.

Area: s = (r 2-r 2) π (r outer circle radius, r inner circle radius) 1 1, arc and bow.

Arc length: l=nπR/ 180=aR (n: arc central angle, a: arc system) Arch area:

Me, the bow is cut on the circle

(1) when the bow is inferior, S bow = S sector -S triangle;

(2) When the arcuate arc is optimal, S-shaped bow = S-shaped sector +S-shaped triangle.

Second, parabolic arch

3/4 of the inscribed triangle with the secant as the base and the tangent point parallel to the base as the vertex II. Three-dimensional graphics

1, ball

Surface area: s = 4 * π * r 2

Volume: v = 4 π r 3/32, cubic

Surface area: s = 6a 2

Volume: v = 33, cuboid

Surface area: S=2(ab+bc+ac)

Volume: V=abc4, prism

Volume: V=Sh (S: bottom area, h: height) 6. Cylindrical surface area: S = 2π RH+π R 2 (R: radius of bottom circle, h: side height).

Volume: V=Sh (S: bottom area, h: height)

=πr^2·h

7. Surface area of cone and pyramid: s = π rh+π r 2 (r: base circle radius, H: side length) Volume of cone and pyramid: V=Sh/3 (S: base area, H: height).

8. prism

Let the areas of the upper and lower bottom surfaces of the truncated cone be S 1 and S2, respectively, the height be h, the volume be v = (1/3) [s1+√ (s1S2)+S2] × h (√ stands for square root) 9, and the volume of the truncated cone be v.

R- bottom radius

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