5 10 12 triangle

Obtuse scalene triangle.

Sides: a = 5   b = 10   c = 12

Area: T = 24.54546022579
Perimeter: p = 27
Semiperimeter: s = 13.5

Angle ∠ A = α = 24.14768479965° = 24°8'49″ = 0.42114420015 rad
Angle ∠ B = β = 54.99003678046° = 54°54'1″ = 0.95881921787 rad
Angle ∠ C = γ = 100.9532784199° = 100°57'10″ = 1.76219584733 rad

Height: ha = 9.81878409032
Height: hb = 4.90989204516
Height: hc = 4.0910767043

Median: ma = 10.75987173957
Median: mb = 7.71436243103
Median: mc = 5.14878150705

Inradius: r = 1.81881186858
Circumradius: R = 6.11113233135

Vertex coordinates: A[12; 0] B[0; 0] C[2.875; 4.0910767043]
Centroid: CG[4.95883333333; 1.36435890143]
Coordinates of the circumscribed circle: U[6; -1.16111514296]
Coordinates of the inscribed circle: I[3.5; 1.81881186858]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155.8533152003° = 155°51'11″ = 0.42114420015 rad
∠ B' = β' = 125.1099632195° = 125°5'59″ = 0.95881921787 rad
∠ C' = γ' = 79.04772158011° = 79°2'50″ = 1.76219584733 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 = 5 ; ; b = 10 ; ; c = 12 ; ;

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

p = a+b+c = 5+10+12 = 27 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 27 }{ 2 } = 13.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 13.5 * (13.5-5)(13.5-10)(13.5-12) } ; ; T = sqrt{ 602.44 } = 24.54 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 24.54 }{ 5 } = 9.82 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 24.54 }{ 10 } = 4.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 24.54 }{ 12 } = 4.09 ; ;

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{ 5**2-10**2-12**2 }{ 2 * 10 * 12 } ) = 24° 8'49" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 10**2-5**2-12**2 }{ 2 * 5 * 12 } ) = 54° 54'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 12**2-5**2-10**2 }{ 2 * 10 * 5 } ) = 100° 57'10" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 24.54 }{ 13.5 } = 1.82 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 24° 8'49" } = 6.11 ; ;




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

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