11 21 25 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 21   c = 25

Area: T = 114.4211097268
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 25.84219327632° = 25°50'31″ = 0.45110268118 rad
Angle ∠ B = β = 56.32105874737° = 56°19'14″ = 0.98329796881 rad
Angle ∠ C = γ = 97.83774797631° = 97°50'15″ = 1.70875861537 rad

Height: ha = 20.80438358669
Height: hb = 10.89772473589
Height: hc = 9.15436877814

Median: ma = 22.42220873248
Median: mb = 16.21095650774
Median: mc = 11.16991539518

Inradius: r = 4.01547753427
Circumradius: R = 12.61878653629

Vertex coordinates: A[25; 0] B[0; 0] C[6.1; 9.15436877814]
Centroid: CG[10.36766666667; 3.05112292605]
Coordinates of the circumscribed circle: U[12.5; -1.7210618004]
Coordinates of the inscribed circle: I[7.5; 4.01547753427]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 154.1588067237° = 154°9'29″ = 0.45110268118 rad
∠ B' = β' = 123.6799412526° = 123°40'46″ = 0.98329796881 rad
∠ C' = γ' = 82.16325202369° = 82°9'45″ = 1.70875861537 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 = 11 ; ; b = 21 ; ; c = 25 ; ;

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

p = a+b+c = 11+21+25 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-11)(28.5-21)(28.5-25) } ; ; T = sqrt{ 13092.19 } = 114.42 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 114.42 }{ 11 } = 20.8 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 114.42 }{ 21 } = 10.9 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 114.42 }{ 25 } = 9.15 ; ;

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{ 11**2-21**2-25**2 }{ 2 * 21 * 25 } ) = 25° 50'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-11**2-25**2 }{ 2 * 11 * 25 } ) = 56° 19'14" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-11**2-21**2 }{ 2 * 21 * 11 } ) = 97° 50'15" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 114.42 }{ 28.5 } = 4.01 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 25° 50'31" } = 12.62 ; ;




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

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