11 25 27 triangle

Acute scalene triangle.

Sides: a = 11   b = 25   c = 27

Area: T = 137.4344302487
Perimeter: p = 63
Semiperimeter: s = 31.5

Angle ∠ A = α = 24.03298638405° = 24°1'47″ = 0.41994002428 rad
Angle ∠ B = β = 67.74113787262° = 67°44'29″ = 1.18223100986 rad
Angle ∠ C = γ = 88.22987574334° = 88°13'44″ = 1.54398823122 rad

Height: ha = 24.98880549976
Height: hb = 10.99547441989
Height: hc = 10.18803187027

Median: ma = 25.43112799521
Median: mb = 16.39435963108
Median: mc = 13.81112273169

Inradius: r = 4.36329937297
Circumradius: R = 13.50664533847

Vertex coordinates: A[27; 0] B[0; 0] C[4.16766666667; 10.18803187027]
Centroid: CG[10.38988888889; 3.39334395676]
Coordinates of the circumscribed circle: U[13.5; 0.41774721955]
Coordinates of the inscribed circle: I[6.5; 4.36329937297]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155.977013616° = 155°58'13″ = 0.41994002428 rad
∠ B' = β' = 112.2598621274° = 112°15'31″ = 1.18223100986 rad
∠ C' = γ' = 91.77112425666° = 91°46'16″ = 1.54398823122 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 = 11 ; ; b = 25 ; ; c = 27 ; ;

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

p = a+b+c = 11+25+27 = 63 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 63 }{ 2 } = 31.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31.5 * (31.5-11)(31.5-25)(31.5-27) } ; ; T = sqrt{ 18888.19 } = 137.43 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 137.43 }{ 11 } = 24.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 137.43 }{ 25 } = 10.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 137.43 }{ 27 } = 10.18 ; ;

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{ 11**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 24° 1'47" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-11**2-27**2 }{ 2 * 11 * 27 } ) = 67° 44'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-11**2-25**2 }{ 2 * 25 * 11 } ) = 88° 13'44" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 137.43 }{ 31.5 } = 4.36 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 24° 1'47" } = 13.51 ; ;




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

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