620 500 620 triangle

Acute isosceles triangle.

Sides: a = 620   b = 500   c = 620

Area: T = 141840.5879525
Perimeter: p = 1740
Semiperimeter: s = 870

Angle ∠ A = α = 66.22200052686° = 66°13'12″ = 1.15657571226 rad
Angle ∠ B = β = 47.56599894628° = 47°33'36″ = 0.83300784083 rad
Angle ∠ C = γ = 66.22200052686° = 66°13'12″ = 1.15657571226 rad

Height: ha = 457.5550256532
Height: hb = 567.36223181
Height: hc = 457.5550256532

Median: ma = 470.213271782
Median: mb = 567.36223181
Median: mc = 470.213271782

Inradius: r = 163.0355148879
Circumradius: R = 338.7610601239

Vertex coordinates: A[620; 0] B[0; 0] C[418.3877096774; 457.5550256532]
Centroid: CG[346.1299032258; 152.5176752177]
Coordinates of the circumscribed circle: U[310; 136.5977016629]
Coordinates of the inscribed circle: I[370; 163.0355148879]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 113.7879994731° = 113°46'48″ = 1.15657571226 rad
∠ B' = β' = 132.4440010537° = 132°26'24″ = 0.83300784083 rad
∠ C' = γ' = 113.7879994731° = 113°46'48″ = 1.15657571226 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 = 620 ; ; b = 500 ; ; c = 620 ; ;

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

p = a+b+c = 620+500+620 = 1740 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1740 }{ 2 } = 870 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 870 * (870-620)(870-500)(870-620) } ; ; T = sqrt{ 20118750000 } = 141840.58 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 141840.58 }{ 620 } = 457.55 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 141840.58 }{ 500 } = 567.36 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 141840.58 }{ 620 } = 457.55 ; ;

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{ 620**2-500**2-620**2 }{ 2 * 500 * 620 } ) = 66° 13'12" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 500**2-620**2-620**2 }{ 2 * 620 * 620 } ) = 47° 33'36" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 620**2-620**2-500**2 }{ 2 * 500 * 620 } ) = 66° 13'12" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 141840.58 }{ 870 } = 163.04 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 620 }{ 2 * sin 66° 13'12" } = 338.76 ; ;




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

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