8 20 27 triangle

Obtuse scalene triangle.

Sides: a = 8   b = 20   c = 27

Area: T = 44.84334777866
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 9.56603836944° = 9°33'37″ = 0.16768601732 rad
Angle ∠ B = β = 24.5333007117° = 24°31'59″ = 0.42881817496 rad
Angle ∠ C = γ = 145.9076609189° = 145°54'24″ = 2.54765507308 rad

Height: ha = 11.21108694467
Height: hb = 4.48443477787
Height: hc = 3.32217390953

Median: ma = 23.42200768573
Median: mb = 17.21991753577
Median: mc = 7.05333679898

Inradius: r = 1.63106719195
Circumradius: R = 24.08437698882

Vertex coordinates: A[27; 0] B[0; 0] C[7.27877777778; 3.32217390953]
Centroid: CG[11.42659259259; 1.10772463651]
Coordinates of the circumscribed circle: U[13.5; -19.94443719387]
Coordinates of the inscribed circle: I[7.5; 1.63106719195]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 170.4439616306° = 170°26'23″ = 0.16768601732 rad
∠ B' = β' = 155.4676992883° = 155°28'1″ = 0.42881817496 rad
∠ C' = γ' = 34.09333908114° = 34°5'36″ = 2.54765507308 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 = 8 ; ; b = 20 ; ; c = 27 ; ;

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

p = a+b+c = 8+20+27 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-8)(27.5-20)(27.5-27) } ; ; T = sqrt{ 2010.94 } = 44.84 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 44.84 }{ 8 } = 11.21 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 44.84 }{ 20 } = 4.48 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 44.84 }{ 27 } = 3.32 ; ;

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{ 8**2-20**2-27**2 }{ 2 * 20 * 27 } ) = 9° 33'37" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-8**2-27**2 }{ 2 * 8 * 27 } ) = 24° 31'59" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-8**2-20**2 }{ 2 * 20 * 8 } ) = 145° 54'24" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 44.84 }{ 27.5 } = 1.63 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 8 }{ 2 * sin 9° 33'37" } = 24.08 ; ;




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

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