14 15 15 triangle

Acute isosceles triangle.

Sides: a = 14   b = 15   c = 15

Area: T = 92.865549413
Perimeter: p = 44
Semiperimeter: s = 22

Angle ∠ A = α = 55.63662785693° = 55°38'11″ = 0.97110362446 rad
Angle ∠ B = β = 62.18218607153° = 62°10'55″ = 1.08552782045 rad
Angle ∠ C = γ = 62.18218607153° = 62°10'55″ = 1.08552782045 rad

Height: ha = 13.26664991614
Height: hb = 12.3822065884
Height: hc = 12.3822065884

Median: ma = 13.26664991614
Median: mb = 12.42197423484
Median: mc = 12.42197423484

Inradius: r = 4.22111588241
Circumradius: R = 8.48800065662

Vertex coordinates: A[15; 0] B[0; 0] C[6.53333333333; 12.3822065884]
Centroid: CG[7.17877777778; 4.12773552947]
Coordinates of the circumscribed circle: U[7.5; 3.95773363976]
Coordinates of the inscribed circle: I[7; 4.22111588241]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.3643721431° = 124°21'49″ = 0.97110362446 rad
∠ B' = β' = 117.8188139285° = 117°49'5″ = 1.08552782045 rad
∠ C' = γ' = 117.8188139285° = 117°49'5″ = 1.08552782045 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 = 14 ; ; b = 15 ; ; c = 15 ; ;

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

p = a+b+c = 14+15+15 = 44 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 44 }{ 2 } = 22 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 22 * (22-14)(22-15)(22-15) } ; ; T = sqrt{ 8624 } = 92.87 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 92.87 }{ 14 } = 13.27 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 92.87 }{ 15 } = 12.38 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 92.87 }{ 15 } = 12.38 ; ;

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{ 14**2-15**2-15**2 }{ 2 * 15 * 15 } ) = 55° 38'11" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-14**2-15**2 }{ 2 * 14 * 15 } ) = 62° 10'55" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 15**2-14**2-15**2 }{ 2 * 15 * 14 } ) = 62° 10'55" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 92.87 }{ 22 } = 4.22 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 14 }{ 2 * sin 55° 38'11" } = 8.48 ; ;




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

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