11 17 22 triangle

Obtuse scalene triangle.

Sides: a = 11   b = 17   c = 22

Area: T = 91.65215138991
Perimeter: p = 50
Semiperimeter: s = 25

Angle ∠ A = α = 29.34881435989° = 29°20'53″ = 0.51222217351 rad
Angle ∠ B = β = 49.24399546573° = 49°14'24″ = 0.85993993323 rad
Angle ∠ C = γ = 101.4121901744° = 101°24'43″ = 1.77699715861 rad

Height: ha = 16.6643911618
Height: hb = 10.7832531047
Height: hc = 8.3321955809

Median: ma = 18.87545860882
Median: mb = 15.17439909055
Median: mc = 9.16551513899

Inradius: r = 3.6666060556
Circumradius: R = 11.22218550054

Vertex coordinates: A[22; 0] B[0; 0] C[7.18218181818; 8.3321955809]
Centroid: CG[9.72772727273; 2.7777318603]
Coordinates of the circumscribed circle: U[11; -2.22203670332]
Coordinates of the inscribed circle: I[8; 3.6666060556]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 150.6521856401° = 150°39'7″ = 0.51222217351 rad
∠ B' = β' = 130.7660045343° = 130°45'36″ = 0.85993993323 rad
∠ C' = γ' = 78.58880982562° = 78°35'17″ = 1.77699715861 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 = 17 ; ; c = 22 ; ;

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

p = a+b+c = 11+17+22 = 50 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 50 }{ 2 } = 25 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25 * (25-11)(25-17)(25-22) } ; ; T = sqrt{ 8400 } = 91.65 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 91.65 }{ 11 } = 16.66 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 91.65 }{ 17 } = 10.78 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 91.65 }{ 22 } = 8.33 ; ;

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**2-17**2-22**2 }{ 2 * 17 * 22 } ) = 29° 20'53" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17**2-11**2-22**2 }{ 2 * 11 * 22 } ) = 49° 14'24" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-11**2-17**2 }{ 2 * 17 * 11 } ) = 101° 24'43" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 91.65 }{ 25 } = 3.67 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 29° 20'53" } = 11.22 ; ;




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

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