11 27 30 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 27   c = 30

Area: T = 147.9732970505
Perimeter: p = 68
Semiperimeter: s = 34

Angle ∠ A = α = 21.43300692627° = 21°25'48″ = 0.37440252676 rad
Angle ∠ B = β = 63.7411340066° = 63°44'29″ = 1.11224962538 rad
Angle ∠ C = γ = 94.82985906713° = 94°49'43″ = 1.65550711322 rad

Height: ha = 26.90441764554
Height: hb = 10.96109607781
Height: hc = 9.86548647003

Median: ma = 28.00444639299
Median: mb = 18.1187670932
Median: mc = 14.14221356237

Inradius: r = 4.35221461913
Circumradius: R = 15.05334249086

Vertex coordinates: A[30; 0] B[0; 0] C[4.86766666667; 9.86548647003]
Centroid: CG[11.62222222222; 3.28882882334]
Coordinates of the circumscribed circle: U[15; -1.26771233088]
Coordinates of the inscribed circle: I[7; 4.35221461913]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 158.5769930737° = 158°34'12″ = 0.37440252676 rad
∠ B' = β' = 116.2598659934° = 116°15'31″ = 1.11224962538 rad
∠ C' = γ' = 85.17114093287° = 85°10'17″ = 1.65550711322 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 = 27 ; ; c = 30 ; ;

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

p = a+b+c = 11+27+30 = 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 * (34-11)(34-27)(34-30) } ; ; T = sqrt{ 21896 } = 147.97 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 147.97 }{ 11 } = 26.9 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 147.97 }{ 27 } = 10.96 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 147.97 }{ 30 } = 9.86 ; ;

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**2-27**2-30**2 }{ 2 * 27 * 30 } ) = 21° 25'48" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-11**2-30**2 }{ 2 * 11 * 30 } ) = 63° 44'29" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-11**2-27**2 }{ 2 * 27 * 11 } ) = 94° 49'43" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 147.97 }{ 34 } = 4.35 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 21° 25'48" } = 15.05 ; ;




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

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