20 25 26 triangle

Acute scalene triangle.

Sides: a = 20   b = 25   c = 26

Area: T = 234.2810681022
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 46.1265829757° = 46°7'33″ = 0.80550475995 rad
Angle ∠ B = β = 64.330033309° = 64°18'1″ = 1.12222525225 rad
Angle ∠ C = γ = 69.5743837153° = 69°34'26″ = 1.21442925316 rad

Height: ha = 23.42880681022
Height: hb = 18.74224544817
Height: hc = 18.02215908478

Median: ma = 23.46327364133
Median: mb = 19.53884236826
Median: mc = 18.53437529929

Inradius: r = 6.59994558034
Circumradius: R = 13.87222492432

Vertex coordinates: A[26; 0] B[0; 0] C[8.67330769231; 18.02215908478]
Centroid: CG[11.55876923077; 6.00771969493]
Coordinates of the circumscribed circle: U[13; 4.84114149859]
Coordinates of the inscribed circle: I[10.5; 6.59994558034]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 133.8744170243° = 133°52'27″ = 0.80550475995 rad
∠ B' = β' = 115.769966691° = 115°41'59″ = 1.12222525225 rad
∠ C' = γ' = 110.4266162847° = 110°25'34″ = 1.21442925316 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 = 25 ; ; c = 26 ; ;

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

p = a+b+c = 20+25+26 = 71 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-20)(35.5-25)(35.5-26) } ; ; T = sqrt{ 54887.44 } = 234.28 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 234.28 }{ 20 } = 23.43 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 234.28 }{ 25 } = 18.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 234.28 }{ 26 } = 18.02 ; ;

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**2-25**2-26**2 }{ 2 * 25 * 26 } ) = 46° 7'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-20**2-26**2 }{ 2 * 20 * 26 } ) = 64° 18'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 26**2-20**2-25**2 }{ 2 * 25 * 20 } ) = 69° 34'26" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 234.28 }{ 35.5 } = 6.6 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 46° 7'33" } = 13.87 ; ;




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

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