11 16 21 triangle

Obtuse scalene triangle.

Sides: a = 11 b = 16 c = 21

Area: T = 86.53332306111
Perimeter: p = 48
Semiperimeter: s = 24

Angle ∠ A = α = 31.00327191339° = 31°10″ = 0.5411099526 rad
Angle ∠ B = β = 48.52215991697° = 48°31'18″ = 0.84768616638 rad
Angle ∠ C = γ = 100.4765681696° = 100°28'32″ = 1.75436314638 rad

Height: h_a = 15.73333146566
Height: h_b = 10.81766538264
Height: h_c = 8.24112600582

Median: m_a = 17.84395627749
Median: m_b = 14.73109198627
Median: m_c = 8.84659030065

Inradius: r = 3.60655512755
Circumradius: R = 10.67879787773

Vertex coordinates: A[21; 0] B[0; 0] C[7.28657142857; 8.24112600582]
Centroid: CG[9.42985714286; 2.74770866861]
Coordinates of the circumscribed circle: U[10.5; -1.94114506868]
Coordinates of the inscribed circle: I[8; 3.60655512755]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.9977280866° = 148°59'50″ = 0.5411099526 rad
∠ B' = β' = 131.478840083° = 131°28'42″ = 0.84768616638 rad
∠ C' = γ' = 79.52443183036° = 79°31'28″ = 1.75436314638 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.

$a = 11 ; ; b = 16 ; ; c = 21 ; ;$

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

$p = a+b+c = 11+16+21 = 48 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 48 }{ 2 } = 24 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 24 * (24-11)(24-16)(24-21) } ; ; T = sqrt{ 7488 } = 86.53 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 86.53 }{ 11 } = 15.73 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 86.53 }{ 16 } = 10.82 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 86.53 }{ 21 } = 8.24 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 11**2-16**2-21**2 }{ 2 * 16 * 21 } ) = 31° 10" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-11**2-21**2 }{ 2 * 11 * 21 } ) = 48° 31'18" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-11**2-16**2 }{ 2 * 16 * 11 } ) = 100° 28'32" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 86.53 }{ 24 } = 3.61 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 31° 10" } = 10.68 ; ;$

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

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