21 21 29 triangle

Acute isosceles triangle.

Sides: a = 21 b = 21 c = 29

Area: T = 220.2621634199
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 46.33221847017° = 46°19'56″ = 0.80986491727 rad
Angle ∠ B = β = 46.33221847017° = 46°19'56″ = 0.80986491727 rad
Angle ∠ C = γ = 87.33656305966° = 87°20'8″ = 1.52442943082 rad

Height: h_a = 20.97772984951
Height: h_b = 20.97772984951
Height: h_c = 15.1990457531

Median: m_a = 23.0388012067
Median: m_b = 23.0388012067
Median: m_c = 15.1990457531

Inradius: r = 6.2054553076
Circumradius: R = 14.51656918118

Vertex coordinates: A[29; 0] B[0; 0] C[14.5; 15.1990457531]
Centroid: CG[14.5; 5.06334858437]
Coordinates of the circumscribed circle: U[14.5; 0.67547657191]
Coordinates of the inscribed circle: I[14.5; 6.2054553076]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 133.6687815298° = 133°40'4″ = 0.80986491727 rad
∠ B' = β' = 133.6687815298° = 133°40'4″ = 0.80986491727 rad
∠ C' = γ' = 92.66443694034° = 92°39'52″ = 1.52442943082 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.

$a = 21 ; ; b = 21 ; ; c = 29 ; ;$

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

$p = a+b+c = 21+21+29 = 71 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-21)(35.5-21)(35.5-29) } ; ; T = sqrt{ 48515.19 } = 220.26 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 220.26 }{ 21 } = 20.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 220.26 }{ 21 } = 20.98 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 220.26 }{ 29 } = 15.19 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 21**2-21**2-29**2 }{ 2 * 21 * 29 } ) = 46° 19'56" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-21**2-29**2 }{ 2 * 21 * 29 } ) = 46° 19'56" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-21**2-21**2 }{ 2 * 21 * 21 } ) = 87° 20'8" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 220.26 }{ 35.5 } = 6.2 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 46° 19'56" } = 14.52 ; ;$

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