14 27 27 triangle

Acute isosceles triangle.

Sides: a = 14   b = 27   c = 27

Area: T = 182.5387667346
Perimeter: p = 68
Semiperimeter: s = 34

Angle ∠ A = α = 30.05222275182° = 30°3'8″ = 0.52545103178 rad
Angle ∠ B = β = 74.97438862409° = 74°58'26″ = 1.30985411679 rad
Angle ∠ C = γ = 74.97438862409° = 74°58'26″ = 1.30985411679 rad

Height: ha = 26.07768096208
Height: hb = 13.52113086923
Height: hc = 13.52113086923

Median: ma = 26.07768096208
Median: mb = 16.74106690428
Median: mc = 16.74106690428

Inradius: r = 5.36987549219
Circumradius: R = 13.97879369217

Vertex coordinates: A[27; 0] B[0; 0] C[3.63296296296; 13.52113086923]
Centroid: CG[10.21098765432; 4.50771028974]
Coordinates of the circumscribed circle: U[13.5; 3.62439095723]
Coordinates of the inscribed circle: I[7; 5.36987549219]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 149.9487772482° = 149°56'52″ = 0.52545103178 rad
∠ B' = β' = 105.0266113759° = 105°1'34″ = 1.30985411679 rad
∠ C' = γ' = 105.0266113759° = 105°1'34″ = 1.30985411679 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 = 14 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 14+27+27 = 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 * (34-14)(34-27)(34-27) } ; ; T = sqrt{ 33320 } = 182.54 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 182.54 }{ 14 } = 26.08 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 182.54 }{ 27 } = 13.52 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 182.54 }{ 27 } = 13.52 ; ;

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{ 14**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 30° 3'8" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-14**2-27**2 }{ 2 * 14 * 27 } ) = 74° 58'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-14**2-27**2 }{ 2 * 27 * 14 } ) = 74° 58'26" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 182.54 }{ 34 } = 5.37 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 30° 3'8" } = 13.98 ; ;




Look also our friend's collection of math examples and problems:

See more informations about triangles or more information about solving triangles.