13 18 21 triangle

Acute scalene triangle.

Sides: a = 13 b = 18 c = 21

Area: T = 116.276553483
Perimeter: p = 52
Semiperimeter: s = 26

Angle ∠ A = α = 37.96875057136° = 37°58'3″ = 0.66326579835 rad
Angle ∠ B = β = 58.41218644948° = 58°24'43″ = 1.01994793577 rad
Angle ∠ C = γ = 83.62106297916° = 83°37'14″ = 1.45994553125 rad

Height: h_a = 17.889854382
Height: h_b = 12.921950387
Height: h_c = 11.074386046

Median: m_a = 18.44658667457
Median: m_b = 14.96766295471
Median: m_c = 11.67326175299

Inradius: r = 4.4722135955
Circumradius: R = 10.56554211937

Vertex coordinates: A[21; 0] B[0; 0] C[6.81095238095; 11.074386046]
Centroid: CG[9.27698412698; 3.691128682]
Coordinates of the circumscribed circle: U[10.5; 1.17439356882]
Coordinates of the inscribed circle: I[8; 4.4722135955]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 142.0322494286° = 142°1'57″ = 0.66326579835 rad
∠ B' = β' = 121.5888135505° = 121°35'17″ = 1.01994793577 rad
∠ C' = γ' = 96.37993702084° = 96°22'46″ = 1.45994553125 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.

$a = 13 ; ; b = 18 ; ; c = 21 ; ;$

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

$p = a+b+c = 13+18+21 = 52 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 52 }{ 2 } = 26 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26 * (26-13)(26-18)(26-21) } ; ; T = sqrt{ 13520 } = 116.28 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 116.28 }{ 13 } = 17.89 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 116.28 }{ 18 } = 12.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 116.28 }{ 21 } = 11.07 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 13**2-18**2-21**2 }{ 2 * 18 * 21 } ) = 37° 58'3" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-13**2-21**2 }{ 2 * 13 * 21 } ) = 58° 24'43" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-13**2-18**2 }{ 2 * 18 * 13 } ) = 83° 37'14" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 116.28 }{ 26 } = 4.47 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 13 }{ 2 * sin 37° 58'3" } = 10.57 ; ;$

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

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