4 4 5 triangle

Acute isosceles triangle.

Sides: a = 4   b = 4   c = 5

Area: T = 7.8066247498
Perimeter: p = 13
Semiperimeter: s = 6.5

Angle ∠ A = α = 51.31878125465° = 51°19'4″ = 0.89656647939 rad
Angle ∠ B = β = 51.31878125465° = 51°19'4″ = 0.89656647939 rad
Angle ∠ C = γ = 77.3644374907° = 77°21'52″ = 1.35502630659 rad

Height: ha = 3.9033123749
Height: hb = 3.9033123749
Height: hc = 3.12224989992

Median: ma = 4.06220192023
Median: mb = 4.06220192023
Median: mc = 3.12224989992

Inradius: r = 1.20109611535
Circumradius: R = 2.56220504609

Vertex coordinates: A[5; 0] B[0; 0] C[2.5; 3.12224989992]
Centroid: CG[2.5; 1.04108329997]
Coordinates of the circumscribed circle: U[2.5; 0.56604485383]
Coordinates of the inscribed circle: I[2.5; 1.20109611535]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 128.6822187453° = 128°40'56″ = 0.89656647939 rad
∠ B' = β' = 128.6822187453° = 128°40'56″ = 0.89656647939 rad
∠ C' = γ' = 102.6365625093° = 102°38'8″ = 1.35502630659 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 = 4 ; ; b = 4 ; ; c = 5 ; ;

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

p = a+b+c = 4+4+5 = 13 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 13 }{ 2 } = 6.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 6.5 * (6.5-4)(6.5-4)(6.5-5) } ; ; T = sqrt{ 60.94 } = 7.81 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 7.81 }{ 4 } = 3.9 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 7.81 }{ 4 } = 3.9 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 7.81 }{ 5 } = 3.12 ; ;

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{ 4**2-4**2-5**2 }{ 2 * 4 * 5 } ) = 51° 19'4" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 4**2-4**2-5**2 }{ 2 * 4 * 5 } ) = 51° 19'4" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 5**2-4**2-4**2 }{ 2 * 4 * 4 } ) = 77° 21'52" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 7.81 }{ 6.5 } = 1.2 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4 }{ 2 * sin 51° 19'4" } = 2.56 ; ;




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

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