12 25 27 triangle

Acute scalene triangle.

Sides: a = 12   b = 25   c = 27

Area: T = 149.6666295471
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 26.32545765375° = 26°19'28″ = 0.45994505348 rad
Angle ∠ B = β = 67.49879771918° = 67°29'53″ = 1.17880619404 rad
Angle ∠ C = γ = 86.17774462707° = 86°10'39″ = 1.50440801784 rad

Height: ha = 24.94443825785
Height: hb = 11.97333036377
Height: hc = 11.08663922571

Median: ma = 25.31879778023
Median: mb = 16.74106690428
Median: mc = 14.22114626533

Inradius: r = 4.67770717335
Circumradius: R = 13.53301003718

Vertex coordinates: A[27; 0] B[0; 0] C[4.59325925926; 11.08663922571]
Centroid: CG[10.53108641975; 3.69554640857]
Coordinates of the circumscribed circle: U[13.5; 0.90220066915]
Coordinates of the inscribed circle: I[7; 4.67770717335]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 153.6755423463° = 153°40'32″ = 0.45994505348 rad
∠ B' = β' = 112.5022022808° = 112°30'7″ = 1.17880619404 rad
∠ C' = γ' = 93.82325537293° = 93°49'21″ = 1.50440801784 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 = 12 ; ; b = 25 ; ; c = 27 ; ;

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

p = a+b+c = 12+25+27 = 64 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 64 }{ 2 } = 32 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-12)(32-25)(32-27) } ; ; T = sqrt{ 22400 } = 149.67 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 149.67 }{ 12 } = 24.94 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 149.67 }{ 25 } = 11.97 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 149.67 }{ 27 } = 11.09 ; ;

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{ 12**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 26° 19'28" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-12**2-27**2 }{ 2 * 12 * 27 } ) = 67° 29'53" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-12**2-25**2 }{ 2 * 25 * 12 } ) = 86° 10'39" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 149.67 }{ 32 } = 4.68 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 12 }{ 2 * sin 26° 19'28" } = 13.53 ; ;




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

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