25 27 27 triangle

Acute isosceles triangle.

Sides: a = 25   b = 27   c = 27

Area: T = 299.1522448594
Perimeter: p = 79
Semiperimeter: s = 39.5

Angle ∠ A = α = 55.15769359562° = 55°9'25″ = 0.96326701377 rad
Angle ∠ B = β = 62.42215320219° = 62°25'18″ = 1.08994612579 rad
Angle ∠ C = γ = 62.42215320219° = 62°25'18″ = 1.08994612579 rad

Height: ha = 23.93221958875
Height: hb = 22.15994406366
Height: hc = 22.15994406366

Median: ma = 23.93221958875
Median: mb = 22.24329764195
Median: mc = 22.24329764195

Inradius: r = 7.57334797112
Circumradius: R = 15.23105288538

Vertex coordinates: A[27; 0] B[0; 0] C[11.57440740741; 22.15994406366]
Centroid: CG[12.85880246914; 7.38664802122]
Coordinates of the circumscribed circle: U[13.5; 7.05111707656]
Coordinates of the inscribed circle: I[12.5; 7.57334797112]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.8433064044° = 124°50'35″ = 0.96326701377 rad
∠ B' = β' = 117.5788467978° = 117°34'42″ = 1.08994612579 rad
∠ C' = γ' = 117.5788467978° = 117°34'42″ = 1.08994612579 rad

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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 = 25 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 25+27+27 = 79 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 79 }{ 2 } = 39.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39.5 * (39.5-25)(39.5-27)(39.5-27) } ; ; T = sqrt{ 89492.19 } = 299.15 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 299.15 }{ 25 } = 23.93 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 299.15 }{ 27 } = 22.16 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 299.15 }{ 27 } = 22.16 ; ;

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{ 25**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 55° 9'25" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 62° 25'18" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-25**2-27**2 }{ 2 * 27 * 25 } ) = 62° 25'18" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 299.15 }{ 39.5 } = 7.57 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 25 }{ 2 * sin 55° 9'25" } = 15.23 ; ;




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