0.5 0.5 0.5 triangle

Equilateral triangle.

Sides: a = 0.5   b = 0.5   c = 0.5

Area: T = 0.10882531755
Perimeter: p = 1.5
Semiperimeter: s = 0.75

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

Height: ha = 0.43330127019
Height: hb = 0.43330127019
Height: hc = 0.43330127019

Median: ma = 0.43330127019
Median: mb = 0.43330127019
Median: mc = 0.43330127019

Inradius: r = 0.14443375673
Circumradius: R = 0.28986751346

Vertex coordinates: A[0.5; 0] B[0; 0] C[0.25; 0.43330127019]
Centroid: CG[0.25; 0.14443375673]
Coordinates of the circumscribed circle: U[0.25; 0.14443375673]
Coordinates of the inscribed circle: I[0.25; 0.14443375673]

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?

a = 0.5 ; ; b = 0.5 ; ; c = 0.5 ; ;

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

p = a+b+c = 0.5+0.5+0.5 = 1.5 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1.5 }{ 2 } = 0.75 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 0.75 * (0.75-0.5)(0.75-0.5)(0.75-0.5) } ; ; T = sqrt{ 0.01 } = 0.11 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 0.11 }{ 0.5 } = 0.43 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 0.11 }{ 0.5 } = 0.43 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 0.11 }{ 0.5 } = 0.43 ; ;

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{ 0.5**2-0.5**2-0.5**2 }{ 2 * 0.5 * 0.5 } ) = 60° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 0.5**2-0.5**2-0.5**2 }{ 2 * 0.5 * 0.5 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.5**2-0.5**2-0.5**2 }{ 2 * 0.5 * 0.5 } ) = 60° ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 0.11 }{ 0.75 } = 0.14 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 0.5 }{ 2 * sin 60° } = 0.29 ; ;




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

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