11 11 17 triangle

Obtuse isosceles triangle.

Sides: a = 11 b = 11 c = 17

Area: T = 59.3488020186
Perimeter: p = 39
Semiperimeter: s = 19.5

Angle ∠ A = α = 39.40105687537° = 39°24'2″ = 0.68876696519 rad
Angle ∠ B = β = 39.40105687537° = 39°24'2″ = 0.68876696519 rad
Angle ∠ C = γ = 101.1998862493° = 101°11'56″ = 1.76662533498 rad

Height: h_a = 10.79105491247
Height: h_b = 10.79105491247
Height: h_c = 6.98221200219

Median: m_a = 13.21993040664
Median: m_b = 13.21993040664
Median: m_c = 6.98221200219

Inradius: r = 3.04334882147
Circumradius: R = 8.66549899759

Vertex coordinates: A[17; 0] B[0; 0] C[8.5; 6.98221200219]
Centroid: CG[8.5; 2.32773733406]
Coordinates of the circumscribed circle: U[8.5; -1.6832869954]
Coordinates of the inscribed circle: I[8.5; 3.04334882147]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140.5999431246° = 140°35'58″ = 0.68876696519 rad
∠ B' = β' = 140.5999431246° = 140°35'58″ = 0.68876696519 rad
∠ C' = γ' = 78.80111375075° = 78°48'4″ = 1.76662533498 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.

$a = 11 ; ; b = 11 ; ; c = 17 ; ;$

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

$p = a+b+c = 11+11+17 = 39 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 39 }{ 2 } = 19.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19.5 * (19.5-11)(19.5-11)(19.5-17) } ; ; T = sqrt{ 3522.19 } = 59.35 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 59.35 }{ 11 } = 10.79 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 59.35 }{ 11 } = 10.79 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 59.35 }{ 17 } = 6.98 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 11**2-11**2-17**2 }{ 2 * 11 * 17 } ) = 39° 24'2" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 11**2-11**2-17**2 }{ 2 * 11 * 17 } ) = 39° 24'2" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 17**2-11**2-11**2 }{ 2 * 11 * 11 } ) = 101° 11'56" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 59.35 }{ 19.5 } = 3.04 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 39° 24'2" } = 8.66 ; ;$

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