21 23 23 triangle

Acute isosceles triangle.

Sides: a = 21   b = 23   c = 23

Area: T = 214.8665510262
Perimeter: p = 67
Semiperimeter: s = 33.5

Angle ∠ A = α = 54.32657768443° = 54°19'33″ = 0.94881636746 rad
Angle ∠ B = β = 62.83771115778° = 62°50'14″ = 1.09767144895 rad
Angle ∠ C = γ = 62.83771115778° = 62°50'14″ = 1.09767144895 rad

Height: ha = 20.46333819297
Height: hb = 18.68439574141
Height: hc = 18.68439574141

Median: ma = 20.46333819297
Median: mb = 18.7821639971
Median: mc = 18.7821639971

Inradius: r = 6.41438958287
Circumradius: R = 12.92655272129

Vertex coordinates: A[23; 0] B[0; 0] C[9.58769565217; 18.68439574141]
Centroid: CG[10.86223188406; 6.22879858047]
Coordinates of the circumscribed circle: U[11.5; 5.90107841624]
Coordinates of the inscribed circle: I[10.5; 6.41438958287]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 125.6744223156° = 125°40'27″ = 0.94881636746 rad
∠ B' = β' = 117.1632888422° = 117°9'46″ = 1.09767144895 rad
∠ C' = γ' = 117.1632888422° = 117°9'46″ = 1.09767144895 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 = 21 ; ; b = 23 ; ; c = 23 ; ;

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

p = a+b+c = 21+23+23 = 67 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 67 }{ 2 } = 33.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33.5 * (33.5-21)(33.5-23)(33.5-23) } ; ; T = sqrt{ 46167.19 } = 214.87 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 214.87 }{ 21 } = 20.46 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 214.87 }{ 23 } = 18.68 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 214.87 }{ 23 } = 18.68 ; ;

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{ 21**2-23**2-23**2 }{ 2 * 23 * 23 } ) = 54° 19'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-21**2-23**2 }{ 2 * 21 * 23 } ) = 62° 50'14" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 23**2-21**2-23**2 }{ 2 * 23 * 21 } ) = 62° 50'14" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 214.87 }{ 33.5 } = 6.41 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 54° 19'33" } = 12.93 ; ;




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

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