9 14 17 triangle

Obtuse scalene triangle.

Sides: a = 9   b = 14   c = 17

Area: T = 62.92985308902
Perimeter: p = 40
Semiperimeter: s = 20

Angle ∠ A = α = 31.9255166456° = 31°55'31″ = 0.55771992689 rad
Angle ∠ B = β = 55.34554309073° = 55°20'44″ = 0.96659599953 rad
Angle ∠ C = γ = 92.72994026368° = 92°43'46″ = 1.61884333894 rad

Height: ha = 13.98441179756
Height: hb = 8.99897901272
Height: hc = 7.40333565753

Median: ma = 14.90880515159
Median: mb = 11.66219037897
Median: mc = 8.1399410298

Inradius: r = 3.14664265445
Circumradius: R = 8.5109653609

Vertex coordinates: A[17; 0] B[0; 0] C[5.11876470588; 7.40333565753]
Centroid: CG[7.37325490196; 2.46877855251]
Coordinates of the circumscribed circle: U[8.5; -0.40552216004]
Coordinates of the inscribed circle: I[6; 3.14664265445]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.0754833544° = 148°4'29″ = 0.55771992689 rad
∠ B' = β' = 124.6554569093° = 124°39'16″ = 0.96659599953 rad
∠ C' = γ' = 87.27105973632° = 87°16'14″ = 1.61884333894 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 = 9 ; ; b = 14 ; ; c = 17 ; ;

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

p = a+b+c = 9+14+17 = 40 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 40 }{ 2 } = 20 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 20 * (20-9)(20-14)(20-17) } ; ; T = sqrt{ 3960 } = 62.93 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 62.93 }{ 9 } = 13.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 62.93 }{ 14 } = 8.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 62.93 }{ 17 } = 7.4 ; ;

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{ 9**2-14**2-17**2 }{ 2 * 14 * 17 } ) = 31° 55'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-9**2-17**2 }{ 2 * 9 * 17 } ) = 55° 20'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 17**2-9**2-14**2 }{ 2 * 14 * 9 } ) = 92° 43'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 62.93 }{ 20 } = 3.15 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 31° 55'31" } = 8.51 ; ;




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

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