21 24 29 triangle

Acute scalene triangle.

Sides: a = 21   b = 24   c = 29

Area: T = 248.1298998708
Perimeter: p = 74
Semiperimeter: s = 37

Angle ∠ A = α = 45.48107045498° = 45°28'51″ = 0.79437880405 rad
Angle ∠ B = β = 54.57548616063° = 54°34'29″ = 0.95325110239 rad
Angle ∠ C = γ = 79.94444338439° = 79°56'40″ = 1.39552935892 rad

Height: ha = 23.63113332103
Height: hb = 20.6777416559
Height: hc = 17.11223447385

Median: ma = 24.45991496173
Median: mb = 22.29334968096
Median: mc = 17.27699160392

Inradius: r = 6.70661891543
Circumradius: R = 14.72662110395

Vertex coordinates: A[29; 0] B[0; 0] C[12.17224137931; 17.11223447385]
Centroid: CG[13.7244137931; 5.70441149128]
Coordinates of the circumscribed circle: U[14.5; 2.57112431974]
Coordinates of the inscribed circle: I[13; 6.70661891543]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 134.519929545° = 134°31'9″ = 0.79437880405 rad
∠ B' = β' = 125.4255138394° = 125°25'31″ = 0.95325110239 rad
∠ C' = γ' = 100.0565566156° = 100°3'20″ = 1.39552935892 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 = 21 ; ; b = 24 ; ; c = 29 ; ;

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

p = a+b+c = 21+24+29 = 74 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 74 }{ 2 } = 37 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 37 * (37-21)(37-24)(37-29) } ; ; T = sqrt{ 61568 } = 248.13 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 248.13 }{ 21 } = 23.63 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 248.13 }{ 24 } = 20.68 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 248.13 }{ 29 } = 17.11 ; ;

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{ 21**2-24**2-29**2 }{ 2 * 24 * 29 } ) = 45° 28'51" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-21**2-29**2 }{ 2 * 21 * 29 } ) = 54° 34'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-21**2-24**2 }{ 2 * 24 * 21 } ) = 79° 56'40" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 248.13 }{ 37 } = 6.71 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 45° 28'51" } = 14.73 ; ;




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

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