20 21 26 triangle

Acute scalene triangle.

Sides: a = 20   b = 21   c = 26

Area: T = 205.9098808699
Perimeter: p = 67
Semiperimeter: s = 33.5

Angle ∠ A = α = 48.95994188107° = 48°57'34″ = 0.85545030581 rad
Angle ∠ B = β = 52.36987700413° = 52°22'8″ = 0.91440074624 rad
Angle ∠ C = γ = 78.67218111479° = 78°40'19″ = 1.3733082133 rad

Height: ha = 20.59108808699
Height: hb = 19.61103627333
Height: hc = 15.83991391307

Median: ma = 21.41326131054
Median: mb = 20.68221178799
Median: mc = 15.85987515272

Inradius: r = 6.1476531603
Circumradius: R = 13.25882963169

Vertex coordinates: A[26; 0] B[0; 0] C[12.21215384615; 15.83991391307]
Centroid: CG[12.73771794872; 5.28797130436]
Coordinates of the circumscribed circle: U[13; 2.60443082051]
Coordinates of the inscribed circle: I[12.5; 6.1476531603]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 131.0410581189° = 131°2'26″ = 0.85545030581 rad
∠ B' = β' = 127.6311229959° = 127°37'52″ = 0.91440074624 rad
∠ C' = γ' = 101.3288188852° = 101°19'41″ = 1.3733082133 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 = 20 ; ; b = 21 ; ; c = 26 ; ;

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

p = a+b+c = 20+21+26 = 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 * (33.5-20)(33.5-21)(33.5-26) } ; ; T = sqrt{ 42398.44 } = 205.91 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 205.91 }{ 20 } = 20.59 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 205.91 }{ 21 } = 19.61 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 205.91 }{ 26 } = 15.84 ; ;

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{ 20**2-21**2-26**2 }{ 2 * 21 * 26 } ) = 48° 57'34" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-20**2-26**2 }{ 2 * 20 * 26 } ) = 52° 22'8" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 26**2-20**2-21**2 }{ 2 * 21 * 20 } ) = 78° 40'19" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 205.91 }{ 33.5 } = 6.15 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 48° 57'34" } = 13.26 ; ;




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

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