13 25 28 triangle

Acute scalene triangle.

Sides: a = 13   b = 25   c = 28

Area: T = 162.4810768093
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 27.66604498993° = 27°39'38″ = 0.48327659233 rad
Angle ∠ B = β = 63.22110584075° = 63°13'16″ = 1.10334156258 rad
Angle ∠ C = γ = 89.11884916932° = 89°7'7″ = 1.55554111045 rad

Height: ha = 24.9977041245
Height: hb = 12.99884614474
Height: hc = 11.60657691495

Median: ma = 25.73442184649
Median: mb = 17.89655301682
Median: mc = 14.17774468788

Inradius: r = 4.92436596392
Circumradius: R = 14.00216570989

Vertex coordinates: A[28; 0] B[0; 0] C[5.85771428571; 11.60657691495]
Centroid: CG[11.28657142857; 3.86985897165]
Coordinates of the circumscribed circle: U[14; 0.21554101092]
Coordinates of the inscribed circle: I[8; 4.92436596392]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 152.3439550101° = 152°20'22″ = 0.48327659233 rad
∠ B' = β' = 116.7798941592° = 116°46'44″ = 1.10334156258 rad
∠ C' = γ' = 90.88215083068° = 90°52'53″ = 1.55554111045 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 = 13 ; ; b = 25 ; ; c = 28 ; ;

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

p = a+b+c = 13+25+28 = 66 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-13)(33-25)(33-28) } ; ; T = sqrt{ 26400 } = 162.48 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 162.48 }{ 13 } = 25 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 162.48 }{ 25 } = 13 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 162.48 }{ 28 } = 11.61 ; ;

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{ 13**2-25**2-28**2 }{ 2 * 25 * 28 } ) = 27° 39'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-13**2-28**2 }{ 2 * 13 * 28 } ) = 63° 13'16" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-13**2-25**2 }{ 2 * 25 * 13 } ) = 89° 7'7" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 162.48 }{ 33 } = 4.92 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 13 }{ 2 * sin 27° 39'38" } = 14 ; ;




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

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