11 24 29 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 24   c = 29

Area: T = 126.9966062931
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 21.40333496394° = 21°24'12″ = 0.37435589222 rad
Angle ∠ B = β = 52.77700302287° = 52°46'12″ = 0.92110107739 rad
Angle ∠ C = γ = 105.8276620132° = 105°49'36″ = 1.84770229576 rad

Height: ha = 23.09901932602
Height: hb = 10.58330052443
Height: hc = 8.75883491677

Median: ma = 26.04332332862
Median: mb = 18.35875597507
Median: mc = 11.75879760163

Inradius: r = 3.96986269666
Circumradius: R = 15.07113333612

Vertex coordinates: A[29; 0] B[0; 0] C[6.65551724138; 8.75883491677]
Centroid: CG[11.88550574713; 2.91994497226]
Coordinates of the circumscribed circle: U[14.5; -4.1110363644]
Coordinates of the inscribed circle: I[8; 3.96986269666]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 158.5976650361° = 158°35'48″ = 0.37435589222 rad
∠ B' = β' = 127.2329969771° = 127°13'48″ = 0.92110107739 rad
∠ C' = γ' = 74.17333798681° = 74°10'24″ = 1.84770229576 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 = 24 ; ; c = 29 ; ;

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

p = a+b+c = 11+24+29 = 64 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 64 }{ 2 } = 32 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-11)(32-24)(32-29) } ; ; T = sqrt{ 16128 } = 127 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 127 }{ 11 } = 23.09 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 127 }{ 24 } = 10.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 127 }{ 29 } = 8.76 ; ;

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**2-24**2-29**2 }{ 2 * 24 * 29 } ) = 21° 24'12" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-11**2-29**2 }{ 2 * 11 * 29 } ) = 52° 46'12" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-11**2-24**2 }{ 2 * 24 * 11 } ) = 105° 49'36" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 127 }{ 32 } = 3.97 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 21° 24'12" } = 15.07 ; ;




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

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