24 24 29 triangle

Acute isosceles triangle.

Sides: a = 24   b = 24   c = 29

Area: T = 277.3076576734
Perimeter: p = 77
Semiperimeter: s = 38.5

Angle ∠ A = α = 52.8311100344° = 52°49'52″ = 0.92220766485 rad
Angle ∠ B = β = 52.8311100344° = 52°49'52″ = 0.92220766485 rad
Angle ∠ C = γ = 74.3387799312° = 74°20'16″ = 1.29774393567 rad

Height: ha = 23.10988813945
Height: hb = 23.10988813945
Height: hc = 19.12545914989

Median: ma = 23.7599208741
Median: mb = 23.7599208741
Median: mc = 19.12545914989

Inradius: r = 7.20327682269
Circumradius: R = 15.05991451857

Vertex coordinates: A[29; 0] B[0; 0] C[14.5; 19.12545914989]
Centroid: CG[14.5; 6.3754863833]
Coordinates of the circumscribed circle: U[14.5; 4.06554463132]
Coordinates of the inscribed circle: I[14.5; 7.20327682269]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 127.1698899656° = 127°10'8″ = 0.92220766485 rad
∠ B' = β' = 127.1698899656° = 127°10'8″ = 0.92220766485 rad
∠ C' = γ' = 105.6622200688° = 105°39'44″ = 1.29774393567 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 = 24 ; ; b = 24 ; ; c = 29 ; ;

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

p = a+b+c = 24+24+29 = 77 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 77 }{ 2 } = 38.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 38.5 * (38.5-24)(38.5-24)(38.5-29) } ; ; T = sqrt{ 76898.94 } = 277.31 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 277.31 }{ 24 } = 23.11 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 277.31 }{ 24 } = 23.11 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 277.31 }{ 29 } = 19.12 ; ;

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{ 24**2-24**2-29**2 }{ 2 * 24 * 29 } ) = 52° 49'52" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-24**2-29**2 }{ 2 * 24 * 29 } ) = 52° 49'52" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-24**2-24**2 }{ 2 * 24 * 24 } ) = 74° 20'16" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 277.31 }{ 38.5 } = 7.2 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 24 }{ 2 * sin 52° 49'52" } = 15.06 ; ;




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

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