11 16 16 triangle

Acute isosceles triangle.

Sides: a = 11   b = 16   c = 16

Area: T = 82.63773856072
Perimeter: p = 43
Semiperimeter: s = 21.5

Angle ∠ A = α = 40.21110195758° = 40°12'40″ = 0.70218146872 rad
Angle ∠ B = β = 69.89444902121° = 69°53'40″ = 1.22198889832 rad
Angle ∠ C = γ = 69.89444902121° = 69°53'40″ = 1.22198889832 rad

Height: ha = 15.02549792013
Height: hb = 10.33296732009
Height: hc = 10.33296732009

Median: ma = 15.02549792013
Median: mb = 11.15879568022
Median: mc = 11.15879568022

Inradius: r = 3.84435993306
Circumradius: R = 8.51991465682

Vertex coordinates: A[16; 0] B[0; 0] C[3.781125; 10.33296732009]
Centroid: CG[6.594375; 3.44332244003]
Coordinates of the circumscribed circle: U[8; 2.92884566328]
Coordinates of the inscribed circle: I[5.5; 3.84435993306]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 139.7898980424° = 139°47'20″ = 0.70218146872 rad
∠ B' = β' = 110.1065509788° = 110°6'20″ = 1.22198889832 rad
∠ C' = γ' = 110.1065509788° = 110°6'20″ = 1.22198889832 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 = 16 ; ;

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

p = a+b+c = 11+16+16 = 43 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 43 }{ 2 } = 21.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 21.5 * (21.5-11)(21.5-16)(21.5-16) } ; ; T = sqrt{ 6828.94 } = 82.64 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 82.64 }{ 11 } = 15.02 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 82.64 }{ 16 } = 10.33 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 82.64 }{ 16 } = 10.33 ; ;

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**2-16**2-16**2 }{ 2 * 16 * 16 } ) = 40° 12'40" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 16**2-11**2-16**2 }{ 2 * 11 * 16 } ) = 69° 53'40" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 16**2-11**2-16**2 }{ 2 * 16 * 11 } ) = 69° 53'40" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 82.64 }{ 21.5 } = 3.84 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 40° 12'40" } = 8.52 ; ;




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

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