10 21 25 triangle

Obtuse scalene triangle.

Sides: a = 10   b = 21   c = 25

Area: T = 102.8798569197
Perimeter: p = 56
Semiperimeter: s = 28

Angle ∠ A = α = 23.07439180656° = 23°4'26″ = 0.40327158416 rad
Angle ∠ B = β = 55.38991229016° = 55°23'21″ = 0.96767225644 rad
Angle ∠ C = γ = 101.5376959033° = 101°32'13″ = 1.77221542476 rad

Height: ha = 20.57657138394
Height: hb = 9.79879589711
Height: hc = 8.23302855358

Median: ma = 22.53988553392
Median: mb = 15.88223801743
Median: mc = 10.68987791632

Inradius: r = 3.67442346142
Circumradius: R = 12.7587759077

Vertex coordinates: A[25; 0] B[0; 0] C[5.68; 8.23302855358]
Centroid: CG[10.22766666667; 2.74334285119]
Coordinates of the circumscribed circle: U[12.5; -2.55215518154]
Coordinates of the inscribed circle: I[7; 3.67442346142]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 156.9266081934° = 156°55'34″ = 0.40327158416 rad
∠ B' = β' = 124.6110877098° = 124°36'39″ = 0.96767225644 rad
∠ C' = γ' = 78.46330409672° = 78°27'47″ = 1.77221542476 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 = 10 ; ; b = 21 ; ; c = 25 ; ;

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

p = a+b+c = 10+21+25 = 56 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 56 }{ 2 } = 28 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28 * (28-10)(28-21)(28-25) } ; ; T = sqrt{ 10584 } = 102.88 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 102.88 }{ 10 } = 20.58 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 102.88 }{ 21 } = 9.8 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 102.88 }{ 25 } = 8.23 ; ;

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{ 10**2-21**2-25**2 }{ 2 * 21 * 25 } ) = 23° 4'26" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-10**2-25**2 }{ 2 * 10 * 25 } ) = 55° 23'21" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-10**2-21**2 }{ 2 * 21 * 10 } ) = 101° 32'13" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 102.88 }{ 28 } = 3.67 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 23° 4'26" } = 12.76 ; ;




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

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