11 19 20 triangle

Acute scalene triangle.

Sides: a = 11   b = 19   c = 20

Area: T = 102.476950766
Perimeter: p = 50
Semiperimeter: s = 25

Angle ∠ A = α = 32.63768975036° = 32°38'13″ = 0.57696213191 rad
Angle ∠ B = β = 68.67663137365° = 68°40'35″ = 1.19986277928 rad
Angle ∠ C = γ = 78.68767887599° = 78°41'12″ = 1.37333435417 rad

Height: ha = 18.63108195745
Height: hb = 10.78662639642
Height: hc = 10.2476950766

Median: ma = 18.71549672722
Median: mb = 13.04879883507
Median: mc = 11.8744342087

Inradius: r = 4.09987803064
Circumradius: R = 10.19881557623

Vertex coordinates: A[20; 0] B[0; 0] C[4; 10.2476950766]
Centroid: CG[8; 3.41656502553]
Coordinates of the circumscribed circle: U[10; 2.00105951495]
Coordinates of the inscribed circle: I[6; 4.09987803064]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 147.3633102496° = 147°21'47″ = 0.57696213191 rad
∠ B' = β' = 111.3243686263° = 111°19'25″ = 1.19986277928 rad
∠ C' = γ' = 101.313321124° = 101°18'48″ = 1.37333435417 rad

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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 = 19 ; ; c = 20 ; ;

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

p = a+b+c = 11+19+20 = 50 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 50 }{ 2 } = 25 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25 * (25-11)(25-19)(25-20) } ; ; T = sqrt{ 10500 } = 102.47 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 102.47 }{ 11 } = 18.63 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 102.47 }{ 19 } = 10.79 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 102.47 }{ 20 } = 10.25 ; ;

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**2-19**2-20**2 }{ 2 * 19 * 20 } ) = 32° 38'13" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 19**2-11**2-20**2 }{ 2 * 11 * 20 } ) = 68° 40'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-11**2-19**2 }{ 2 * 19 * 11 } ) = 78° 41'12" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 102.47 }{ 25 } = 4.1 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 32° 38'13" } = 10.2 ; ;




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