9 23 23 triangle

Acute isosceles triangle.

Sides: a = 9   b = 23   c = 23

Area: T = 101.5499692118
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 22.56656485759° = 22°33'56″ = 0.39438448655 rad
Angle ∠ B = β = 78.7177175712° = 78°43'2″ = 1.3743873894 rad
Angle ∠ C = γ = 78.7177175712° = 78°43'2″ = 1.3743873894 rad

Height: ha = 22.55554871373
Height: hb = 8.82660601842
Height: hc = 8.82660601842

Median: ma = 22.55554871373
Median: mb = 13.14334394281
Median: mc = 13.14334394281

Inradius: r = 3.69108978952
Circumradius: R = 11.72766365559

Vertex coordinates: A[23; 0] B[0; 0] C[1.76108695652; 8.82660601842]
Centroid: CG[8.25436231884; 2.94220200614]
Coordinates of the circumscribed circle: U[11.5; 2.29443419348]
Coordinates of the inscribed circle: I[4.5; 3.69108978952]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 157.4344351424° = 157°26'4″ = 0.39438448655 rad
∠ B' = β' = 101.2832824288° = 101°16'58″ = 1.3743873894 rad
∠ C' = γ' = 101.2832824288° = 101°16'58″ = 1.3743873894 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 = 23 ; ; c = 23 ; ;

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

p = a+b+c = 9+23+23 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-9)(27.5-23)(27.5-23) } ; ; T = sqrt{ 10302.19 } = 101.5 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 101.5 }{ 9 } = 22.56 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 101.5 }{ 23 } = 8.83 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 101.5 }{ 23 } = 8.83 ; ;

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**2-23**2-23**2 }{ 2 * 23 * 23 } ) = 22° 33'56" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-9**2-23**2 }{ 2 * 9 * 23 } ) = 78° 43'2" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-9**2-23**2 }{ 2 * 23 * 9 } ) = 78° 43'2" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 101.5 }{ 27.5 } = 3.69 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 22° 33'56" } = 11.73 ; ;




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

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