14 22 29 triangle

Obtuse scalene triangle.

Sides: a = 14   b = 22   c = 29

Area: T = 148.6477023179
Perimeter: p = 65
Semiperimeter: s = 32.5

Angle ∠ A = α = 27.77435432671° = 27°46'25″ = 0.4854739775 rad
Angle ∠ B = β = 47.07554658418° = 47°4'32″ = 0.82216218758 rad
Angle ∠ C = γ = 105.1510990891° = 105°9'4″ = 1.83552310028 rad

Height: ha = 21.23552890256
Height: hb = 13.51333657436
Height: hc = 10.25215188399

Median: ma = 24.7698932153
Median: mb = 19.93774020374
Median: mc = 11.39107857499

Inradius: r = 4.57437545594
Circumradius: R = 15.0222164267

Vertex coordinates: A[29; 0] B[0; 0] C[9.53444827586; 10.25215188399]
Centroid: CG[12.84548275862; 3.41771729466]
Coordinates of the circumscribed circle: U[14.5; -3.92662474789]
Coordinates of the inscribed circle: I[10.5; 4.57437545594]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 152.2266456733° = 152°13'35″ = 0.4854739775 rad
∠ B' = β' = 132.9254534158° = 132°55'28″ = 0.82216218758 rad
∠ C' = γ' = 74.84990091089° = 74°50'56″ = 1.83552310028 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 = 14 ; ; b = 22 ; ; c = 29 ; ;

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

p = a+b+c = 14+22+29 = 65 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 65 }{ 2 } = 32.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32.5 * (32.5-14)(32.5-22)(32.5-29) } ; ; T = sqrt{ 22095.94 } = 148.65 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 148.65 }{ 14 } = 21.24 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 148.65 }{ 22 } = 13.51 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 148.65 }{ 29 } = 10.25 ; ;

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{ 14**2-22**2-29**2 }{ 2 * 22 * 29 } ) = 27° 46'25" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-14**2-29**2 }{ 2 * 14 * 29 } ) = 47° 4'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-14**2-22**2 }{ 2 * 22 * 14 } ) = 105° 9'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 148.65 }{ 32.5 } = 4.57 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 27° 46'25" } = 15.02 ; ;




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

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