6 15 18 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 15   c = 18

Area: T = 42.15437364892
Perimeter: p = 39
Semiperimeter: s = 19.5

Angle ∠ A = α = 18.19548723388° = 18°11'42″ = 0.31875604293 rad
Angle ∠ B = β = 51.31878125465° = 51°19'4″ = 0.89656647939 rad
Angle ∠ C = γ = 110.4877315115° = 110°29'14″ = 1.92883674304 rad

Height: ha = 14.05112454964
Height: hb = 5.62204981986
Height: hc = 4.68437484988

Median: ma = 16.29441707368
Median: mb = 11.12442977306
Median: mc = 7.03656236397

Inradius: r = 2.16217300764
Circumradius: R = 9.60876892283

Vertex coordinates: A[18; 0] B[0; 0] C[3.75; 4.68437484988]
Centroid: CG[7.25; 1.56112494996]
Coordinates of the circumscribed circle: U[9; -3.36326912299]
Coordinates of the inscribed circle: I[4.5; 2.16217300764]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 161.8055127661° = 161°48'18″ = 0.31875604293 rad
∠ B' = β' = 128.6822187453° = 128°40'56″ = 0.89656647939 rad
∠ C' = γ' = 69.51326848853° = 69°30'46″ = 1.92883674304 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 = 15 ; ; c = 18 ; ;

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

p = a+b+c = 6+15+18 = 39 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 39 }{ 2 } = 19.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19.5 * (19.5-6)(19.5-15)(19.5-18) } ; ; T = sqrt{ 1776.94 } = 42.15 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 42.15 }{ 6 } = 14.05 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 42.15 }{ 15 } = 5.62 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 42.15 }{ 18 } = 4.68 ; ;

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**2-15**2-18**2 }{ 2 * 15 * 18 } ) = 18° 11'42" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-6**2-18**2 }{ 2 * 6 * 18 } ) = 51° 19'4" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 18**2-6**2-15**2 }{ 2 * 15 * 6 } ) = 110° 29'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 42.15 }{ 19.5 } = 2.16 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 18° 11'42" } = 9.61 ; ;




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

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