4 29 29 triangle

Acute isosceles triangle.

Sides: a = 4   b = 29   c = 29

Area: T = 57.8621904566
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 7.90991442442° = 7°54'33″ = 0.13880406081 rad
Angle ∠ B = β = 86.04554278779° = 86°2'44″ = 1.50217760228 rad
Angle ∠ C = γ = 86.04554278779° = 86°2'44″ = 1.50217760228 rad

Height: ha = 28.9310952283
Height: hb = 3.9990476177
Height: hc = 3.9990476177

Median: ma = 28.9310952283
Median: mb = 14.77332867027
Median: mc = 14.77332867027

Inradius: r = 1.86765130505
Circumradius: R = 14.53546062545

Vertex coordinates: A[29; 0] B[0; 0] C[0.2765862069; 3.9990476177]
Centroid: CG[9.75986206897; 1.33301587257]
Coordinates of the circumscribed circle: U[14.5; 1.00223866382]
Coordinates of the inscribed circle: I[2; 1.86765130505]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 172.0910855756° = 172°5'27″ = 0.13880406081 rad
∠ B' = β' = 93.95545721221° = 93°57'16″ = 1.50217760228 rad
∠ C' = γ' = 93.95545721221° = 93°57'16″ = 1.50217760228 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 = 4 ; ; b = 29 ; ; c = 29 ; ;

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

p = a+b+c = 4+29+29 = 62 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 62 }{ 2 } = 31 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31 * (31-4)(31-29)(31-29) } ; ; T = sqrt{ 3348 } = 57.86 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 57.86 }{ 4 } = 28.93 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 57.86 }{ 29 } = 3.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 57.86 }{ 29 } = 3.99 ; ;

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{ 4**2-29**2-29**2 }{ 2 * 29 * 29 } ) = 7° 54'33" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-4**2-29**2 }{ 2 * 4 * 29 } ) = 86° 2'44" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-4**2-29**2 }{ 2 * 29 * 4 } ) = 86° 2'44" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 57.86 }{ 31 } = 1.87 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4 }{ 2 * sin 7° 54'33" } = 14.53 ; ;




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

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