1 14 14 triangle

Acute isosceles triangle.

Sides: a = 1   b = 14   c = 14

Area: T = 6.99655342898
Perimeter: p = 29
Semiperimeter: s = 14.5

Angle ∠ A = α = 4.09334261954° = 4°5'36″ = 0.07114437648 rad
Angle ∠ B = β = 87.95332869023° = 87°57'12″ = 1.53550744444 rad
Angle ∠ C = γ = 87.95332869023° = 87°57'12″ = 1.53550744444 rad

Height: ha = 13.99110685796
Height: hb = 0.99993620414
Height: hc = 0.99993620414

Median: ma = 13.99110685796
Median: mb = 7.03656236397
Median: mc = 7.03656236397

Inradius: r = 0.48224506407
Circumradius: R = 7.00444685609

Vertex coordinates: A[14; 0] B[0; 0] C[0.03657142857; 0.99993620414]
Centroid: CG[4.67985714286; 0.33331206805]
Coordinates of the circumscribed circle: U[7; 0.25501595915]
Coordinates of the inscribed circle: I[0.5; 0.48224506407]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 175.9076573805° = 175°54'24″ = 0.07114437648 rad
∠ B' = β' = 92.04767130977° = 92°2'48″ = 1.53550744444 rad
∠ C' = γ' = 92.04767130977° = 92°2'48″ = 1.53550744444 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 = 1 ; ; b = 14 ; ; c = 14 ; ;

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

p = a+b+c = 1+14+14 = 29 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 29 }{ 2 } = 14.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 14.5 * (14.5-1)(14.5-14)(14.5-14) } ; ; T = sqrt{ 48.94 } = 7 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 7 }{ 1 } = 13.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 7 }{ 14 } = 1 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 7 }{ 14 } = 1 ; ;

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{ 1**2-14**2-14**2 }{ 2 * 14 * 14 } ) = 4° 5'36" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-1**2-14**2 }{ 2 * 1 * 14 } ) = 87° 57'12" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 14**2-1**2-14**2 }{ 2 * 14 * 1 } ) = 87° 57'12" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 7 }{ 14.5 } = 0.48 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 1 }{ 2 * sin 4° 5'36" } = 7 ; ;




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

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