14 15 18 triangle

Acute scalene triangle.

Sides: a = 14   b = 15   c = 18

Area: T = 102.1611330747
Perimeter: p = 47
Semiperimeter: s = 23.5

Angle ∠ A = α = 49.17985691625° = 49°10'43″ = 0.85883279533 rad
Angle ∠ B = β = 54.17545797066° = 54°10'28″ = 0.94655247868 rad
Angle ∠ C = γ = 76.64768511309° = 76°38'49″ = 1.33877399135 rad

Height: ha = 14.5944475821
Height: hb = 13.62215107663
Height: hc = 11.35112589719

Median: ma = 15.01766574177
Median: mb = 14.27441024236
Median: mc = 11.38798066767

Inradius: r = 4.34772906701
Circumradius: R = 9.25500752789

Vertex coordinates: A[18; 0] B[0; 0] C[8.19444444444; 11.35112589719]
Centroid: CG[8.73114814815; 3.78437529906]
Coordinates of the circumscribed circle: U[9; 2.13663269096]
Coordinates of the inscribed circle: I[8.5; 4.34772906701]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 130.8211430838° = 130°49'17″ = 0.85883279533 rad
∠ B' = β' = 125.8255420293° = 125°49'32″ = 0.94655247868 rad
∠ C' = γ' = 103.3533148869° = 103°21'11″ = 1.33877399135 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 = 14 ; ; b = 15 ; ; c = 18 ; ;

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

p = a+b+c = 14+15+18 = 47 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 47 }{ 2 } = 23.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 23.5 * (23.5-14)(23.5-15)(23.5-18) } ; ; T = sqrt{ 10436.94 } = 102.16 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 102.16 }{ 14 } = 14.59 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 102.16 }{ 15 } = 13.62 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 102.16 }{ 18 } = 11.35 ; ;

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{ 14**2-15**2-18**2 }{ 2 * 15 * 18 } ) = 49° 10'43" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-14**2-18**2 }{ 2 * 14 * 18 } ) = 54° 10'28" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 18**2-14**2-15**2 }{ 2 * 15 * 14 } ) = 76° 38'49" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 102.16 }{ 23.5 } = 4.35 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 49° 10'43" } = 9.25 ; ;




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

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