13 20 22 triangle

Acute scalene triangle.

Sides: a = 13   b = 20   c = 22

Area: T = 128.251146198
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 35.65990876961° = 35°39'33″ = 0.62223684886 rad
Angle ∠ B = β = 63.7498786065° = 63°44'56″ = 1.1132626211 rad
Angle ∠ C = γ = 80.59221262388° = 80°35'32″ = 1.40765979541 rad

Height: ha = 19.73109941508
Height: hb = 12.8255146198
Height: hc = 11.65992238164

Median: ma = 19.99437490231
Median: mb = 15.05499169433
Median: mc = 12.78767118525

Inradius: r = 4.66436895265
Circumradius: R = 11.15499703623

Vertex coordinates: A[22; 0] B[0; 0] C[5.75; 11.65992238164]
Centroid: CG[9.25; 3.88664079388]
Coordinates of the circumscribed circle: U[11; 1.82325913092]
Coordinates of the inscribed circle: I[7.5; 4.66436895265]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 144.3410912304° = 144°20'27″ = 0.62223684886 rad
∠ B' = β' = 116.2511213935° = 116°15'4″ = 1.1132626211 rad
∠ C' = γ' = 99.40878737612° = 99°24'28″ = 1.40765979541 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 = 20 ; ; c = 22 ; ;

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

p = a+b+c = 13+20+22 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-13)(27.5-20)(27.5-22) } ; ; T = sqrt{ 16448.44 } = 128.25 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 128.25 }{ 13 } = 19.73 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 128.25 }{ 20 } = 12.83 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 128.25 }{ 22 } = 11.66 ; ;

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**2-20**2-22**2 }{ 2 * 20 * 22 } ) = 35° 39'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-13**2-22**2 }{ 2 * 13 * 22 } ) = 63° 44'56" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-13**2-20**2 }{ 2 * 20 * 13 } ) = 80° 35'32" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 128.25 }{ 27.5 } = 4.66 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 13 }{ 2 * sin 35° 39'33" } = 11.15 ; ;




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

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