2.5 2.5 4.5 triangle

Obtuse isosceles triangle.

Sides: a = 2.5   b = 2.5   c = 4.5

Area: T = 2.45218806557
Perimeter: p = 9.5
Semiperimeter: s = 4.75

Angle ∠ A = α = 25.84219327632° = 25°50'31″ = 0.45110268118 rad
Angle ∠ B = β = 25.84219327632° = 25°50'31″ = 0.45110268118 rad
Angle ∠ C = γ = 128.3166134474° = 128°18'58″ = 2.243953903 rad

Height: ha = 1.96215045246
Height: hb = 1.96215045246
Height: hc = 1.09897247359

Median: ma = 3.41986985828
Median: mb = 3.41986985828
Median: mc = 1.09897247359

Inradius: r = 0.51661854012
Circumradius: R = 2.86876966734

Vertex coordinates: A[4.5; 0] B[0; 0] C[2.25; 1.09897247359]
Centroid: CG[2.25; 0.36332415786]
Coordinates of the circumscribed circle: U[2.25; -1.77879719375]
Coordinates of the inscribed circle: I[2.25; 0.51661854012]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 154.1588067237° = 154°9'29″ = 0.45110268118 rad
∠ B' = β' = 154.1588067237° = 154°9'29″ = 0.45110268118 rad
∠ C' = γ' = 51.68438655263° = 51°41'2″ = 2.243953903 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 = 2.5 ; ; b = 2.5 ; ; c = 4.5 ; ;

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

p = a+b+c = 2.5+2.5+4.5 = 9.5 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 9.5 }{ 2 } = 4.75 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.75 * (4.75-2.5)(4.75-2.5)(4.75-4.5) } ; ; T = sqrt{ 6.01 } = 2.45 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 2.45 }{ 2.5 } = 1.96 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 2.45 }{ 2.5 } = 1.96 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 2.45 }{ 4.5 } = 1.09 ; ;

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{ 2.5**2-2.5**2-4.5**2 }{ 2 * 2.5 * 4.5 } ) = 25° 50'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 2.5**2-2.5**2-4.5**2 }{ 2 * 2.5 * 4.5 } ) = 25° 50'31" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 4.5**2-2.5**2-2.5**2 }{ 2 * 2.5 * 2.5 } ) = 128° 18'58" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 2.45 }{ 4.75 } = 0.52 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.5 }{ 2 * sin 25° 50'31" } = 2.87 ; ;




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

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