17 20 24 triangle

Acute scalene triangle.

Sides: a = 17   b = 20   c = 24

Area: T = 167.6366325121
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 44.30655563826° = 44°18'20″ = 0.77332778358 rad
Angle ∠ B = β = 55.26600284682° = 55°15'36″ = 0.96444694415 rad
Angle ∠ C = γ = 80.43444151492° = 80°26'4″ = 1.40438453763 rad

Height: ha = 19.72219206024
Height: hb = 16.76436325121
Height: hc = 13.97696937601

Median: ma = 20.39899485041
Median: mb = 18.23545825288
Median: mc = 14.16598022585

Inradius: r = 5.49662729548
Circumradius: R = 12.16992001929

Vertex coordinates: A[24; 0] B[0; 0] C[9.68875; 13.97696937601]
Centroid: CG[11.22991666667; 4.65765645867]
Coordinates of the circumscribed circle: U[12; 2.02222347379]
Coordinates of the inscribed circle: I[10.5; 5.49662729548]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135.6944443617° = 135°41'40″ = 0.77332778358 rad
∠ B' = β' = 124.7439971532° = 124°44'24″ = 0.96444694415 rad
∠ C' = γ' = 99.56655848508° = 99°33'56″ = 1.40438453763 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 = 17 ; ; b = 20 ; ; c = 24 ; ;

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

p = a+b+c = 17+20+24 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-17)(30.5-20)(30.5-24) } ; ; T = sqrt{ 28101.94 } = 167.64 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 167.64 }{ 17 } = 19.72 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 167.64 }{ 20 } = 16.76 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 167.64 }{ 24 } = 13.97 ; ;

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{ 17**2-20**2-24**2 }{ 2 * 20 * 24 } ) = 44° 18'20" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-17**2-24**2 }{ 2 * 17 * 24 } ) = 55° 15'36" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 24**2-17**2-20**2 }{ 2 * 20 * 17 } ) = 80° 26'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 167.64 }{ 30.5 } = 5.5 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 44° 18'20" } = 12.17 ; ;




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

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