6 24 29 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 24   c = 29

Area: T = 43.66327701824
Perimeter: p = 59
Semiperimeter: s = 29.5

Angle ∠ A = α = 7.2087767416° = 7°12'28″ = 0.12657992731 rad
Angle ∠ B = β = 30.12438565658° = 30°7'26″ = 0.52657604805 rad
Angle ∠ C = γ = 142.6688376018° = 142°40'6″ = 2.49900329 rad

Height: ha = 14.55442567275
Height: hb = 3.63985641819
Height: hc = 3.01112255298

Median: ma = 26.44880623109
Median: mb = 17.16110023017
Median: mc = 9.78551928954

Inradius: r = 1.48800939045
Circumradius: R = 23.91105305421

Vertex coordinates: A[29; 0] B[0; 0] C[5.19896551724; 3.01112255298]
Centroid: CG[11.39765517241; 1.00437418433]
Coordinates of the circumscribed circle: U[14.5; -19.0122192688]
Coordinates of the inscribed circle: I[5.5; 1.48800939045]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 172.7922232584° = 172°47'32″ = 0.12657992731 rad
∠ B' = β' = 149.8766143434° = 149°52'34″ = 0.52657604805 rad
∠ C' = γ' = 37.33216239818° = 37°19'54″ = 2.49900329 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 = 24 ; ; c = 29 ; ;

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

p = a+b+c = 6+24+29 = 59 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 59 }{ 2 } = 29.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29.5 * (29.5-6)(29.5-24)(29.5-29) } ; ; T = sqrt{ 1906.44 } = 43.66 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 43.66 }{ 6 } = 14.55 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 43.66 }{ 24 } = 3.64 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 43.66 }{ 29 } = 3.01 ; ;

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**2-24**2-29**2 }{ 2 * 24 * 29 } ) = 7° 12'28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-6**2-29**2 }{ 2 * 6 * 29 } ) = 30° 7'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-6**2-24**2 }{ 2 * 24 * 6 } ) = 142° 40'6" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 43.66 }{ 29.5 } = 1.48 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 7° 12'28" } = 23.91 ; ;




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

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