14 14 23 triangle

Obtuse isosceles triangle.

Sides: a = 14   b = 14   c = 23

Area: T = 91.82201366804
Perimeter: p = 51
Semiperimeter: s = 25.5

Angle ∠ A = α = 34.77219440319° = 34°46'19″ = 0.60768849107 rad
Angle ∠ B = β = 34.77219440319° = 34°46'19″ = 0.60768849107 rad
Angle ∠ C = γ = 110.4566111936° = 110°27'22″ = 1.92878228322 rad

Height: ha = 13.11771623829
Height: hb = 13.11771623829
Height: hc = 7.98443597113

Median: ma = 17.70659312096
Median: mb = 17.70659312096
Median: mc = 7.98443597113

Inradius: r = 3.60107896737
Circumradius: R = 12.27439961053

Vertex coordinates: A[23; 0] B[0; 0] C[11.5; 7.98443597113]
Centroid: CG[11.5; 2.66114532371]
Coordinates of the circumscribed circle: U[11.5; -4.29896363939]
Coordinates of the inscribed circle: I[11.5; 3.60107896737]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 145.2288055968° = 145°13'41″ = 0.60768849107 rad
∠ B' = β' = 145.2288055968° = 145°13'41″ = 0.60768849107 rad
∠ C' = γ' = 69.54438880639° = 69°32'38″ = 1.92878228322 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 = 14 ; ; c = 23 ; ;

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

p = a+b+c = 14+14+23 = 51 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 51 }{ 2 } = 25.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25.5 * (25.5-14)(25.5-14)(25.5-23) } ; ; T = sqrt{ 8430.94 } = 91.82 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 91.82 }{ 14 } = 13.12 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 91.82 }{ 14 } = 13.12 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 91.82 }{ 23 } = 7.98 ; ;

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**2-14**2-23**2 }{ 2 * 14 * 23 } ) = 34° 46'19" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-14**2-23**2 }{ 2 * 14 * 23 } ) = 34° 46'19" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-14**2-14**2 }{ 2 * 14 * 14 } ) = 110° 27'22" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 91.82 }{ 25.5 } = 3.6 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 34° 46'19" } = 12.27 ; ;




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

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