20 22 24 triangle

Acute scalene triangle.

Sides: a = 20   b = 22   c = 24

Area: T = 206.0854933947
Perimeter: p = 66
Semiperimeter: s = 33

Angle ∠ A = α = 51.31878125465° = 51°19'4″ = 0.89656647939 rad
Angle ∠ B = β = 59.17695025682° = 59°10'10″ = 1.03327026366 rad
Angle ∠ C = γ = 69.51326848853° = 69°30'46″ = 1.21332252231 rad

Height: ha = 20.60884933947
Height: hb = 18.73549939952
Height: hc = 17.17437444956

Median: ma = 20.73664413533
Median: mb = 19.15772440607
Median: mc = 17.26326765016

Inradius: r = 6.24549979984
Circumradius: R = 12.81102523044

Vertex coordinates: A[24; 0] B[0; 0] C[10.25; 17.17437444956]
Centroid: CG[11.41766666667; 5.72545814985]
Coordinates of the circumscribed circle: U[12; 4.48435883065]
Coordinates of the inscribed circle: I[11; 6.24549979984]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 128.6822187453° = 128°40'56″ = 0.89656647939 rad
∠ B' = β' = 120.8330497432° = 120°49'50″ = 1.03327026366 rad
∠ C' = γ' = 110.4877315115° = 110°29'14″ = 1.21332252231 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 = 20 ; ; b = 22 ; ; c = 24 ; ;

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

p = a+b+c = 20+22+24 = 66 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 66 }{ 2 } = 33 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 33 * (33-20)(33-22)(33-24) } ; ; T = sqrt{ 42471 } = 206.08 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 206.08 }{ 20 } = 20.61 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 206.08 }{ 22 } = 18.73 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 206.08 }{ 24 } = 17.17 ; ;

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{ 20**2-22**2-24**2 }{ 2 * 22 * 24 } ) = 51° 19'4" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-20**2-24**2 }{ 2 * 20 * 24 } ) = 59° 10'10" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 24**2-20**2-22**2 }{ 2 * 22 * 20 } ) = 69° 30'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 206.08 }{ 33 } = 6.24 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 51° 19'4" } = 12.81 ; ;




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

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