9 12 16 triangle

Obtuse scalene triangle.

Sides: a = 9 b = 12 c = 16

Area: T = 53.44109721094
Perimeter: p = 37
Semiperimeter: s = 18.5

Angle ∠ A = α = 33.82662859376° = 33°49'35″ = 0.59903800633 rad
Angle ∠ B = β = 47.92222147824° = 47°55'20″ = 0.83664004328 rad
Angle ∠ C = γ = 98.251149928° = 98°15'5″ = 1.71548121575 rad

Height: h_a = 11.87657715799
Height: h_b = 8.90768286849
Height: h_c = 6.68801215137

Median: m_a = 13.40770876778
Median: m_b = 11.51108644332
Median: m_c = 6.96441941386

Inradius: r = 2.88987011951
Circumradius: R = 8.08436852877

Vertex coordinates: A[16; 0] B[0; 0] C[6.031125; 6.68801215137]
Centroid: CG[7.344375; 2.22767071712]
Coordinates of the circumscribed circle: U[8; -1.16601585367]
Coordinates of the inscribed circle: I[6.5; 2.88987011951]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.1743714062° = 146°10'25″ = 0.59903800633 rad
∠ B' = β' = 132.0787785218° = 132°4'40″ = 0.83664004328 rad
∠ C' = γ' = 81.749850072° = 81°44'55″ = 1.71548121575 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.

$a = 9 ; ; b = 12 ; ; c = 16 ; ;$

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

$p = a+b+c = 9+12+16 = 37 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 37 }{ 2 } = 18.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 18.5 * (18.5-9)(18.5-12)(18.5-16) } ; ; T = sqrt{ 2855.94 } = 53.44 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 53.44 }{ 9 } = 11.88 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 53.44 }{ 12 } = 8.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 53.44 }{ 16 } = 6.68 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 9**2-12**2-16**2 }{ 2 * 12 * 16 } ) = 33° 49'35" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 12**2-9**2-16**2 }{ 2 * 9 * 16 } ) = 47° 55'20" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 16**2-9**2-12**2 }{ 2 * 12 * 9 } ) = 98° 15'5" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 53.44 }{ 18.5 } = 2.89 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 33° 49'35" } = 8.08 ; ;$

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

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