40 40 56.57 triangle

Obtuse isosceles triangle.

Sides: a = 40   b = 40   c = 56.57

Area: T = 8009.999998938
Perimeter: p = 136.57
Semiperimeter: s = 68.285

Angle ∠ A = α = 44.99985237384° = 44°59'55″ = 0.78553723978 rad
Angle ∠ B = β = 44.99985237384° = 44°59'55″ = 0.78553723978 rad
Angle ∠ C = γ = 90.00329525231° = 90°11″ = 1.5710847858 rad

Height: ha = 409.9999999469
Height: hb = 409.9999999469
Height: hc = 28.28435424761

Median: ma = 44.72222813595
Median: mb = 44.72222813595
Median: mc = 28.28435424761

Inradius: r = 11.71656037041
Circumradius: R = 28.28550000376

Vertex coordinates: A[56.57; 0] B[0; 0] C[28.285; 28.28435424761]
Centroid: CG[28.285; 9.4287847492]
Coordinates of the circumscribed circle: U[28.285; -0.00114575614]
Coordinates of the inscribed circle: I[28.285; 11.71656037041]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135.0011476262° = 135°5″ = 0.78553723978 rad
∠ B' = β' = 135.0011476262° = 135°5″ = 0.78553723978 rad
∠ C' = γ' = 89.99770474769° = 89°59'49″ = 1.5710847858 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 = 40 ; ; b = 40 ; ; c = 56.57 ; ;

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

p = a+b+c = 40+40+56.57 = 136.57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 136.57 }{ 2 } = 68.29 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 68.29 * (68.29-40)(68.29-40)(68.29-56.57) } ; ; T = sqrt{ 640000 } = 800 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 800 }{ 40 } = 40 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 800 }{ 40 } = 40 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 800 }{ 56.57 } = 28.28 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 40**2+56.57**2-40**2 }{ 2 * 40 * 56.57 } ) = 44° 59'55" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 40**2+56.57**2-40**2 }{ 2 * 40 * 56.57 } ) = 44° 59'55" ; ;
 gamma = 180° - alpha - beta = 180° - 44° 59'55" - 44° 59'55" = 90° 11" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 800 }{ 68.29 } = 11.72 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 40 }{ 2 * sin 44° 59'55" } = 28.29 ; ;

8. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 40**2+2 * 56.57**2 - 40**2 } }{ 2 } = 44.722 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 56.57**2+2 * 40**2 - 40**2 } }{ 2 } = 44.722 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 40**2+2 * 40**2 - 56.57**2 } }{ 2 } = 28.284 ; ;
Calculate another triangle


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

See more information about triangles or more details on solving triangles.