17 22 22 triangle

Acute isosceles triangle.

Sides: a = 17 b = 22 c = 22

Area: T = 172.4798803046
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 45.4576845195° = 45°27'25″ = 0.79333716162 rad
Angle ∠ B = β = 67.27215774025° = 67°16'18″ = 1.17441105187 rad
Angle ∠ C = γ = 67.27215774025° = 67°16'18″ = 1.17441105187 rad

Height: h_a = 20.29216238877
Height: h_b = 15.6879891186
Height: h_c = 15.6879891186

Median: m_a = 20.29216238877
Median: m_b = 16.29441707368
Median: m_c = 16.29441707368

Inradius: r = 5.65550427228
Circumradius: R = 11.92661031714

Vertex coordinates: A[22; 0] B[0; 0] C[6.56881818182; 15.6879891186]
Centroid: CG[9.52327272727; 5.22766303953]
Coordinates of the circumscribed circle: U[11; 4.6087812589]
Coordinates of the inscribed circle: I[8.5; 5.65550427228]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 134.5433154805° = 134°32'35″ = 0.79333716162 rad
∠ B' = β' = 112.7288422598° = 112°43'42″ = 1.17441105187 rad
∠ C' = γ' = 112.7288422598° = 112°43'42″ = 1.17441105187 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.

$a = 17 ; ; b = 22 ; ; c = 22 ; ;$

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

$p = a+b+c = 17+22+22 = 61 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-17)(30.5-22)(30.5-22) } ; ; T = sqrt{ 29748.94 } = 172.48 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 172.48 }{ 17 } = 20.29 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 172.48 }{ 22 } = 15.68 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 172.48 }{ 22 } = 15.68 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 17**2-22**2-22**2 }{ 2 * 22 * 22 } ) = 45° 27'25" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-17**2-22**2 }{ 2 * 17 * 22 } ) = 67° 16'18" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-17**2-22**2 }{ 2 * 22 * 17 } ) = 67° 16'18" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 172.48 }{ 30.5 } = 5.66 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 45° 27'25" } = 11.93 ; ;$

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