5 5 5 triangle

Equilateral triangle.

Sides: a = 5   b = 5   c = 5

Area: T = 10.82553175473
Perimeter: p = 15
Semiperimeter: s = 7.5

Angle ∠ A = α = 60° = 1.04771975512 rad
Angle ∠ B = β = 60° = 1.04771975512 rad
Angle ∠ C = γ = 60° = 1.04771975512 rad

Height: ha = 4.33301270189
Height: hb = 4.33301270189
Height: hc = 4.33301270189

Median: ma = 4.33301270189
Median: mb = 4.33301270189
Median: mc = 4.33301270189

Inradius: r = 1.4433375673
Circumradius: R = 2.88767513459

Vertex coordinates: A[5; 0] B[0; 0] C[2.5; 4.33301270189]
Centroid: CG[2.5; 1.4433375673]
Coordinates of the circumscribed circle: U[2.5; 1.4433375673]
Coordinates of the inscribed circle: I[2.5; 1.4433375673]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 120° = 1.04771975512 rad
∠ B' = β' = 120° = 1.04771975512 rad
∠ C' = γ' = 120° = 1.04771975512 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 = 5 ; ; b = 5 ; ; c = 5 ; ;

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

p = a+b+c = 5+5+5 = 15 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 15 }{ 2 } = 7.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 7.5 * (7.5-5)(7.5-5)(7.5-5) } ; ; T = sqrt{ 117.19 } = 10.83 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 10.83 }{ 5 } = 4.33 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 10.83 }{ 5 } = 4.33 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 10.83 }{ 5 } = 4.33 ; ;

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{ 5**2-5**2-5**2 }{ 2 * 5 * 5 } ) = 60° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 5**2-5**2-5**2 }{ 2 * 5 * 5 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 5**2-5**2-5**2 }{ 2 * 5 * 5 } ) = 60° ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 10.83 }{ 7.5 } = 1.44 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 60° } = 2.89 ; ;




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

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