11 14 14 triangle

Acute isosceles triangle.

Sides: a = 11   b = 14   c = 14

Area: T = 70.80991625427
Perimeter: p = 39
Semiperimeter: s = 19.5

Angle ∠ A = α = 46.26547927986° = 46°15'53″ = 0.80774729621 rad
Angle ∠ B = β = 66.86876036007° = 66°52'3″ = 1.16770598458 rad
Angle ∠ C = γ = 66.86876036007° = 66°52'3″ = 1.16770598458 rad

Height: ha = 12.87443931896
Height: hb = 10.1165594649
Height: hc = 10.1165594649

Median: ma = 12.87443931896
Median: mb = 10.46442247682
Median: mc = 10.46442247682

Inradius: r = 3.63112391048
Circumradius: R = 7.61220092463

Vertex coordinates: A[14; 0] B[0; 0] C[4.32114285714; 10.1165594649]
Centroid: CG[6.10771428571; 3.3721864883]
Coordinates of the circumscribed circle: U[7; 2.99904322039]
Coordinates of the inscribed circle: I[5.5; 3.63112391048]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 133.7355207201° = 133°44'7″ = 0.80774729621 rad
∠ B' = β' = 113.1322396399° = 113°7'57″ = 1.16770598458 rad
∠ C' = γ' = 113.1322396399° = 113°7'57″ = 1.16770598458 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 = 14 ; ; c = 14 ; ;

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

p = a+b+c = 11+14+14 = 39 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 39 }{ 2 } = 19.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 19.5 * (19.5-11)(19.5-14)(19.5-14) } ; ; T = sqrt{ 5013.94 } = 70.81 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 70.81 }{ 11 } = 12.87 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 70.81 }{ 14 } = 10.12 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 70.81 }{ 14 } = 10.12 ; ;

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**2-14**2-14**2 }{ 2 * 14 * 14 } ) = 46° 15'53" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 14**2-11**2-14**2 }{ 2 * 11 * 14 } ) = 66° 52'3" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 14**2-11**2-14**2 }{ 2 * 14 * 11 } ) = 66° 52'3" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 70.81 }{ 19.5 } = 3.63 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 46° 15'53" } = 7.61 ; ;




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

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