20 24 24 triangle

Acute isosceles triangle.

Sides: a = 20 b = 24 c = 24

Area: T = 218.1744242293
Perimeter: p = 68
Semiperimeter: s = 34

Angle ∠ A = α = 49.24986367043° = 49°14'55″ = 0.86595508626 rad
Angle ∠ B = β = 65.37656816478° = 65°22'32″ = 1.14110208955 rad
Angle ∠ C = γ = 65.37656816478° = 65°22'32″ = 1.14110208955 rad

Height: h_a = 21.81774242293
Height: h_b = 18.18111868577
Height: h_c = 18.18111868577

Median: m_a = 21.81774242293
Median: m_b = 18.5477236991
Median: m_c = 18.5477236991

Inradius: r = 6.41768894792
Circumradius: R = 13.22004583572

Vertex coordinates: A[24; 0] B[0; 0] C[8.33333333333; 18.18111868577]
Centroid: CG[10.77877777778; 6.06603956192]
Coordinates of the circumscribed circle: U[12; 5.55001909822]
Coordinates of the inscribed circle: I[10; 6.41768894792]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 130.7511363296° = 130°45'5″ = 0.86595508626 rad
∠ B' = β' = 114.6244318352° = 114°37'28″ = 1.14110208955 rad
∠ C' = γ' = 114.6244318352° = 114°37'28″ = 1.14110208955 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.

$a = 20 ; ; b = 24 ; ; c = 24 ; ;$

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

$p = a+b+c = 20+24+24 = 68 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 68 }{ 2 } = 34 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 34 * (34-20)(34-24)(34-24) } ; ; T = sqrt{ 47600 } = 218.17 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 218.17 }{ 20 } = 21.82 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 218.17 }{ 24 } = 18.18 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 218.17 }{ 24 } = 18.18 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 20**2-24**2-24**2 }{ 2 * 24 * 24 } ) = 49° 14'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-20**2-24**2 }{ 2 * 20 * 24 } ) = 65° 22'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 24**2-20**2-24**2 }{ 2 * 24 * 20 } ) = 65° 22'32" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 218.17 }{ 34 } = 6.42 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 49° 14'55" } = 13.2 ; ;$

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