21 22 24 triangle

Acute scalene triangle.

Sides: a = 21   b = 22   c = 24

Area: T = 213.8898843795
Perimeter: p = 67
Semiperimeter: s = 33.5

Angle ∠ A = α = 54.11440117604° = 54°6'50″ = 0.94444676767 rad
Angle ∠ B = β = 58.07876244183° = 58°4'39″ = 1.01436457678 rad
Angle ∠ C = γ = 67.80883638213° = 67°48'30″ = 1.18334792091 rad

Height: ha = 20.37703660757
Height: hb = 19.4444440345
Height: hc = 17.82440703163

Median: ma = 20.48878012485
Median: mb = 19.6855019685
Median: mc = 17.84765682976

Inradius: r = 6.38547416058
Circumradius: R = 12.96600027323

Vertex coordinates: A[24; 0] B[0; 0] C[11.10441666667; 17.82440703163]
Centroid: CG[11.70113888889; 5.94113567721]
Coordinates of the circumscribed circle: U[12; 4.89550659671]
Coordinates of the inscribed circle: I[11.5; 6.38547416058]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 125.886598824° = 125°53'10″ = 0.94444676767 rad
∠ B' = β' = 121.9222375582° = 121°55'21″ = 1.01436457678 rad
∠ C' = γ' = 112.1921636179° = 112°11'30″ = 1.18334792091 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 = 21 ; ; b = 22 ; ; c = 24 ; ;

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

p = a+b+c = 21+22+24 = 67 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 67 }{ 2 } = 33.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33.5 * (33.5-21)(33.5-22)(33.5-24) } ; ; T = sqrt{ 45748.44 } = 213.89 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 213.89 }{ 21 } = 20.37 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 213.89 }{ 22 } = 19.44 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 213.89 }{ 24 } = 17.82 ; ;

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{ 21**2-22**2-24**2 }{ 2 * 22 * 24 } ) = 54° 6'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-21**2-24**2 }{ 2 * 21 * 24 } ) = 58° 4'39" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 24**2-21**2-22**2 }{ 2 * 22 * 21 } ) = 67° 48'30" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 213.89 }{ 33.5 } = 6.38 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 54° 6'50" } = 12.96 ; ;




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

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