6 21 22 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 21   c = 22

Area: T = 62.97656897541
Perimeter: p = 49
Semiperimeter: s = 24.5

Angle ∠ A = α = 15.82203528829° = 15°49'13″ = 0.27661172466 rad
Angle ∠ B = β = 72.58878929405° = 72°35'16″ = 1.26768977289 rad
Angle ∠ C = γ = 91.59217541766° = 91°35'30″ = 1.59985776781 rad

Height: ha = 20.99218965847
Height: hb = 5.99876847385
Height: hc = 5.72550627049

Median: ma = 21.2965539439
Median: mb = 12.23772382505
Median: mc = 10.84397416943

Inradius: r = 2.57704363165
Circumradius: R = 11.00442462847

Vertex coordinates: A[22; 0] B[0; 0] C[1.79554545455; 5.72550627049]
Centroid: CG[7.93218181818; 1.9088354235]
Coordinates of the circumscribed circle: U[11; -0.30656735079]
Coordinates of the inscribed circle: I[3.5; 2.57704363165]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 164.1879647117° = 164°10'47″ = 0.27661172466 rad
∠ B' = β' = 107.412210706° = 107°24'44″ = 1.26768977289 rad
∠ C' = γ' = 88.40882458234° = 88°24'30″ = 1.59985776781 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 = 6 ; ; b = 21 ; ; c = 22 ; ;

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

p = a+b+c = 6+21+22 = 49 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 49 }{ 2 } = 24.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 24.5 * (24.5-6)(24.5-21)(24.5-22) } ; ; T = sqrt{ 3965.94 } = 62.98 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 62.98 }{ 6 } = 20.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 62.98 }{ 21 } = 6 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 62.98 }{ 22 } = 5.73 ; ;

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{ 6**2-21**2-22**2 }{ 2 * 21 * 22 } ) = 15° 49'13" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-6**2-22**2 }{ 2 * 6 * 22 } ) = 72° 35'16" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-6**2-21**2 }{ 2 * 21 * 6 } ) = 91° 35'30" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 62.98 }{ 24.5 } = 2.57 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 15° 49'13" } = 11 ; ;




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

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