6 6 10 triangle

Obtuse isosceles triangle.

Sides: a = 6   b = 6   c = 10

Area: T = 16.58331239518
Perimeter: p = 22
Semiperimeter: s = 11

Angle ∠ A = α = 33.55773097619° = 33°33'26″ = 0.58656855435 rad
Angle ∠ B = β = 33.55773097619° = 33°33'26″ = 0.58656855435 rad
Angle ∠ C = γ = 112.8855380476° = 112°53'7″ = 1.97702215667 rad

Height: ha = 5.52877079839
Height: hb = 5.52877079839
Height: hc = 3.31766247904

Median: ma = 7.68111457479
Median: mb = 7.68111457479
Median: mc = 3.31766247904

Inradius: r = 1.50875567229
Circumradius: R = 5.42772042024

Vertex coordinates: A[10; 0] B[0; 0] C[5; 3.31766247904]
Centroid: CG[5; 1.10655415968]
Coordinates of the circumscribed circle: U[5; -2.1110579412]
Coordinates of the inscribed circle: I[5; 1.50875567229]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 146.4432690238° = 146°26'34″ = 0.58656855435 rad
∠ B' = β' = 146.4432690238° = 146°26'34″ = 0.58656855435 rad
∠ C' = γ' = 67.11546195238° = 67°6'53″ = 1.97702215667 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 = 6 ; ; b = 6 ; ; c = 10 ; ;

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

p = a+b+c = 6+6+10 = 22 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 22 }{ 2 } = 11 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 11 * (11-6)(11-6)(11-10) } ; ; T = sqrt{ 275 } = 16.58 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 16.58 }{ 6 } = 5.53 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 16.58 }{ 6 } = 5.53 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 16.58 }{ 10 } = 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{ 6**2-6**2-10**2 }{ 2 * 6 * 10 } ) = 33° 33'26" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 6**2-6**2-10**2 }{ 2 * 6 * 10 } ) = 33° 33'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 10**2-6**2-6**2 }{ 2 * 6 * 6 } ) = 112° 53'7" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 16.58 }{ 11 } = 1.51 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 33° 33'26" } = 5.43 ; ;




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

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