Equilateral triangle calculator (a)

Please enter one property of the equilateral triangle

Use symbols: a, h, T, p, r, R


You have entered side a.

Equilateral triangle.

Sides: a = 9   b = 9   c = 9

Area: T = 35.07440288533
Perimeter: p = 27
Semiperimeter: s = 13.5

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

Height: ha = 7.79442286341
Height: hb = 7.79442286341
Height: hc = 7.79442286341

Median: ma = 7.79442286341
Median: mb = 7.79442286341
Median: mc = 7.79442286341

Inradius: r = 2.59880762114
Circumradius: R = 5.19661524227

Vertex coordinates: A[9; 0] B[0; 0] C[4.5; 7.79442286341]
Centroid: CG[4.5; 2.59880762114]
Coordinates of the circumscribed circle: U[4.5; 2.59880762114]
Coordinates of the inscribed circle: I[4.5; 2.59880762114]

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

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How did we calculate this triangle?

1. Input data entered: side a

a = 9 ; ;

2. From side a we calculate b,c:

b = c = a = 9 ; ;


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 = 9 ; ; b = 9 ; ; c = 9 ; ;

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

p = a+b+c = 9+9+9 = 27 ; ;

4. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 27 }{ 2 } = 13.5 ; ;

5. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 13.5 * (13.5-9)(13.5-9)(13.5-9) } ; ; T = sqrt{ 1230.19 } = 35.07 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 35.07 }{ 9 } = 7.79 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 35.07 }{ 9 } = 7.79 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 35.07 }{ 9 } = 7.79 ; ;

7. 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{ 9**2-9**2-9**2 }{ 2 * 9 * 9 } ) = 60° ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 9**2-9**2-9**2 }{ 2 * 9 * 9 } ) = 60° ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 9**2-9**2-9**2 }{ 2 * 9 * 9 } ) = 60° ; ;

8. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 35.07 }{ 13.5 } = 2.6 ; ;

9. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 60° } = 5.2 ; ;




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