20 20 29 triangle

Obtuse isosceles triangle.

Sides: a = 20   b = 20   c = 29

Area: T = 199.7377171052
Perimeter: p = 69
Semiperimeter: s = 34.5

Angle ∠ A = α = 43.53111521674° = 43°31'52″ = 0.76597619325 rad
Angle ∠ B = β = 43.53111521674° = 43°31'52″ = 0.76597619325 rad
Angle ∠ C = γ = 92.93876956653° = 92°56'16″ = 1.62220687886 rad

Height: ha = 19.97437171052
Height: hb = 19.97437171052
Height: hc = 13.7754977314

Median: ma = 22.81444690931
Median: mb = 22.81444690931
Median: mc = 13.7754977314

Inradius: r = 5.78994832189
Circumradius: R = 14.51990801728

Vertex coordinates: A[29; 0] B[0; 0] C[14.5; 13.7754977314]
Centroid: CG[14.5; 4.59216591047]
Coordinates of the circumscribed circle: U[14.5; -0.74441028589]
Coordinates of the inscribed circle: I[14.5; 5.78994832189]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.4698847833° = 136°28'8″ = 0.76597619325 rad
∠ B' = β' = 136.4698847833° = 136°28'8″ = 0.76597619325 rad
∠ C' = γ' = 87.06223043347° = 87°3'44″ = 1.62220687886 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 = 20 ; ; c = 29 ; ;

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

p = a+b+c = 20+20+29 = 69 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 69 }{ 2 } = 34.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 34.5 * (34.5-20)(34.5-20)(34.5-29) } ; ; T = sqrt{ 39894.94 } = 199.74 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 199.74 }{ 20 } = 19.97 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 199.74 }{ 20 } = 19.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 199.74 }{ 29 } = 13.77 ; ;

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**2-20**2-29**2 }{ 2 * 20 * 29 } ) = 43° 31'52" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-20**2-29**2 }{ 2 * 20 * 29 } ) = 43° 31'52" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-20**2-20**2 }{ 2 * 20 * 20 } ) = 92° 56'16" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 199.74 }{ 34.5 } = 5.79 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 43° 31'52" } = 14.52 ; ;




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

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