6 22 25 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 22   c = 25

Area: T = 60.55552433733
Perimeter: p = 53
Semiperimeter: s = 26.5

Angle ∠ A = α = 12.72108322063° = 12°43'15″ = 0.22220204056 rad
Angle ∠ B = β = 53.8432991799° = 53°50'35″ = 0.9439737486 rad
Angle ∠ C = γ = 113.4366175995° = 113°26'10″ = 1.9879834762 rad

Height: ha = 20.18550811244
Height: hb = 5.50550221248
Height: hc = 4.84444194699

Median: ma = 23.35659414283
Median: mb = 14.47441148261
Median: mc = 10.18657743937

Inradius: r = 2.28551035235
Circumradius: R = 13.62439234465

Vertex coordinates: A[25; 0] B[0; 0] C[3.54; 4.84444194699]
Centroid: CG[9.51333333333; 1.615480649]
Coordinates of the circumscribed circle: U[12.5; -5.41986059162]
Coordinates of the inscribed circle: I[4.5; 2.28551035235]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 167.2799167794° = 167°16'45″ = 0.22220204056 rad
∠ B' = β' = 126.1577008201° = 126°9'25″ = 0.9439737486 rad
∠ C' = γ' = 66.56438240053° = 66°33'50″ = 1.9879834762 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 = 6 ; ; b = 22 ; ; c = 25 ; ;

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

p = a+b+c = 6+22+25 = 53 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 53 }{ 2 } = 26.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26.5 * (26.5-6)(26.5-22)(26.5-25) } ; ; T = sqrt{ 3666.94 } = 60.56 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 60.56 }{ 6 } = 20.19 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 60.56 }{ 22 } = 5.51 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 60.56 }{ 25 } = 4.84 ; ;

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{ 6**2-22**2-25**2 }{ 2 * 22 * 25 } ) = 12° 43'15" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-6**2-25**2 }{ 2 * 6 * 25 } ) = 53° 50'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-6**2-22**2 }{ 2 * 22 * 6 } ) = 113° 26'10" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 60.56 }{ 26.5 } = 2.29 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 12° 43'15" } = 13.62 ; ;




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

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