6 14 14 triangle

Acute isosceles triangle.

Sides: a = 6   b = 14   c = 14

Area: T = 41.02443829935
Perimeter: p = 34
Semiperimeter: s = 17

Angle ∠ A = α = 24.74772502324° = 24°44'50″ = 0.43219209974 rad
Angle ∠ B = β = 77.62663748838° = 77°37'35″ = 1.35548358281 rad
Angle ∠ C = γ = 77.62663748838° = 77°37'35″ = 1.35548358281 rad

Height: ha = 13.67547943312
Height: hb = 5.86106261419
Height: hc = 5.86106261419

Median: ma = 13.67547943312
Median: mb = 8.18553527719
Median: mc = 8.18553527719

Inradius: r = 2.41331989996
Circumradius: R = 7.16664697564

Vertex coordinates: A[14; 0] B[0; 0] C[1.28657142857; 5.86106261419]
Centroid: CG[5.09552380952; 1.95435420473]
Coordinates of the circumscribed circle: U[7; 1.53656720907]
Coordinates of the inscribed circle: I[3; 2.41331989996]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155.2532749768° = 155°15'10″ = 0.43219209974 rad
∠ B' = β' = 102.3743625116° = 102°22'25″ = 1.35548358281 rad
∠ C' = γ' = 102.3743625116° = 102°22'25″ = 1.35548358281 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 = 6 ; ; b = 14 ; ; c = 14 ; ;

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

p = a+b+c = 6+14+14 = 34 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 34 }{ 2 } = 17 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 17 * (17-6)(17-14)(17-14) } ; ; T = sqrt{ 1683 } = 41.02 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 41.02 }{ 6 } = 13.67 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 41.02 }{ 14 } = 5.86 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 41.02 }{ 14 } = 5.86 ; ;

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{ 6**2-14**2-14**2 }{ 2 * 14 * 14 } ) = 24° 44'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-6**2-14**2 }{ 2 * 6 * 14 } ) = 77° 37'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 14**2-6**2-14**2 }{ 2 * 14 * 6 } ) = 77° 37'35" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 41.02 }{ 17 } = 2.41 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 24° 44'50" } = 7.17 ; ;




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

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