12 20 22 triangle

Acute scalene triangle.

Sides: a = 12   b = 20   c = 22

Area: T = 119.0598808998
Perimeter: p = 54
Semiperimeter: s = 27

Angle ∠ A = α = 32.76437577589° = 32°45'50″ = 0.57218354482 rad
Angle ∠ B = β = 64.41769980226° = 64°25'1″ = 1.12442887097 rad
Angle ∠ C = γ = 82.81992442185° = 82°49'9″ = 1.44554684956 rad

Height: ha = 19.8433134833
Height: hb = 11.90658808998
Height: hc = 10.82435280907

Median: ma = 20.14994416796
Median: mb = 14.62987388383
Median: mc = 12.28882057274

Inradius: r = 4.41095855184
Circumradius: R = 11.08769578749

Vertex coordinates: A[22; 0] B[0; 0] C[5.18218181818; 10.82435280907]
Centroid: CG[9.06106060606; 3.60878426969]
Coordinates of the circumscribed circle: U[11; 1.38658697344]
Coordinates of the inscribed circle: I[7; 4.41095855184]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.2366242241° = 147°14'10″ = 0.57218354482 rad
∠ B' = β' = 115.5833001977° = 115°34'59″ = 1.12442887097 rad
∠ C' = γ' = 97.18107557815° = 97°10'51″ = 1.44554684956 rad

Calculate another triangle




How did we calculate this triangle?

Now we know the lengths of all three sides of the triangle and the triangle is uniquely determined. Next we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.

a = 12 ; ; b = 20 ; ; c = 22 ; ;

1. The triangle circumference is the sum of the lengths of its three sides

p = a+b+c = 12+20+22 = 54 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 54 }{ 2 } = 27 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27 * (27-12)(27-20)(27-22) } ; ; T = sqrt{ 14175 } = 119.06 ; ;

4. Calculate the heights of the triangle from its area.

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 119.06 }{ 12 } = 19.84 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 119.06 }{ 20 } = 11.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 119.06 }{ 22 } = 10.82 ; ;

5. Calculation of the inner angles of the triangle using a Law of Cosines

a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 12**2-20**2-22**2 }{ 2 * 20 * 22 } ) = 32° 45'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-12**2-22**2 }{ 2 * 12 * 22 } ) = 64° 25'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-12**2-20**2 }{ 2 * 20 * 12 } ) = 82° 49'9" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 119.06 }{ 27 } = 4.41 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 12 }{ 2 * sin 32° 45'50" } = 11.09 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.