11 26 29 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 26   c = 29

Area: T = 142.5766295365
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 22.22113929634° = 22°13'17″ = 0.38878364716 rad
Angle ∠ B = β = 63.36768812508° = 63°22'1″ = 1.10659607145 rad
Angle ∠ C = γ = 94.41217257858° = 94°24'42″ = 1.64877954675 rad

Height: ha = 25.92329627936
Height: hb = 10.96774073358
Height: hc = 9.83328479562

Median: ma = 26.9866107537
Median: mb = 17.66435217327
Median: mc = 13.7220422734

Inradius: r = 4.32204937989
Circumradius: R = 14.54330907339

Vertex coordinates: A[29; 0] B[0; 0] C[4.93110344828; 9.83328479562]
Centroid: CG[11.31103448276; 3.27876159854]
Coordinates of the circumscribed circle: U[14.5; -1.11986992872]
Coordinates of the inscribed circle: I[7; 4.32204937989]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 157.7798607037° = 157°46'43″ = 0.38878364716 rad
∠ B' = β' = 116.6333118749° = 116°37'59″ = 1.10659607145 rad
∠ C' = γ' = 85.58882742142° = 85°35'18″ = 1.64877954675 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 = 11 ; ; b = 26 ; ; c = 29 ; ;

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

p = a+b+c = 11+26+29 = 66 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-11)(33-26)(33-29) } ; ; T = sqrt{ 20328 } = 142.58 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 142.58 }{ 11 } = 25.92 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 142.58 }{ 26 } = 10.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 142.58 }{ 29 } = 9.83 ; ;

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{ 11**2-26**2-29**2 }{ 2 * 26 * 29 } ) = 22° 13'17" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-11**2-29**2 }{ 2 * 11 * 29 } ) = 63° 22'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-11**2-26**2 }{ 2 * 26 * 11 } ) = 94° 24'42" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 142.58 }{ 33 } = 4.32 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 22° 13'17" } = 14.54 ; ;




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

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