25 26 27 triangle

Acute scalene triangle.

Sides: a = 25   b = 26   c = 27

Area: T = 291.8499276168
Perimeter: p = 78
Semiperimeter: s = 39

Angle ∠ A = α = 56.25110114041° = 56°15'4″ = 0.98217653566 rad
Angle ∠ B = β = 59.85328697332° = 59°51'10″ = 1.04546296436 rad
Angle ∠ C = γ = 63.89661188627° = 63°53'46″ = 1.11551976534 rad

Height: ha = 23.34879420935
Height: hb = 22.45499443206
Height: hc = 21.61884649014

Median: ma = 23.3721991785
Median: mb = 22.53988553392
Median: mc = 21.63990850084

Inradius: r = 7.48333147735
Circumradius: R = 15.03334448576

Vertex coordinates: A[27; 0] B[0; 0] C[12.55655555556; 21.61884649014]
Centroid: CG[13.18551851852; 7.20661549671]
Coordinates of the circumscribed circle: U[13.5; 6.61547157373]
Coordinates of the inscribed circle: I[13; 7.48333147735]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 123.7498988596° = 123°44'56″ = 0.98217653566 rad
∠ B' = β' = 120.1477130267° = 120°8'50″ = 1.04546296436 rad
∠ C' = γ' = 116.1043881137° = 116°6'14″ = 1.11551976534 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 = 25 ; ; b = 26 ; ; c = 27 ; ;

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

p = a+b+c = 25+26+27 = 78 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 78 }{ 2 } = 39 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39 * (39-25)(39-26)(39-27) } ; ; T = sqrt{ 85176 } = 291.85 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 291.85 }{ 25 } = 23.35 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 291.85 }{ 26 } = 22.45 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 291.85 }{ 27 } = 21.62 ; ;

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**2-26**2-27**2 }{ 2 * 26 * 27 } ) = 56° 15'4" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 26**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 59° 51'10" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-25**2-26**2 }{ 2 * 26 * 25 } ) = 63° 53'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 291.85 }{ 39 } = 7.48 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 25 }{ 2 * sin 56° 15'4" } = 15.03 ; ;




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

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