1 27 27 triangle

Acute isosceles triangle.

Sides: a = 1   b = 27   c = 27

Area: T = 13.49876849867
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 2.12221872153° = 2°7'20″ = 0.03770391542 rad
Angle ∠ B = β = 88.93989063924° = 88°56'20″ = 1.55222767497 rad
Angle ∠ C = γ = 88.93989063924° = 88°56'20″ = 1.55222767497 rad

Height: ha = 26.99553699734
Height: hb = 10.9998285175
Height: hc = 10.9998285175

Median: ma = 26.99553699734
Median: mb = 13.51985058346
Median: mc = 13.51985058346

Inradius: r = 0.49108249086
Circumradius: R = 13.50223154104

Vertex coordinates: A[27; 0] B[0; 0] C[0.01985185185; 10.9998285175]
Centroid: CG[9.00661728395; 0.33332761725]
Coordinates of the circumscribed circle: U[13.5; 0.2550042878]
Coordinates of the inscribed circle: I[0.5; 0.49108249086]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 177.8787812785° = 177°52'40″ = 0.03770391542 rad
∠ B' = β' = 91.06110936076° = 91°3'40″ = 1.55222767497 rad
∠ C' = γ' = 91.06110936076° = 91°3'40″ = 1.55222767497 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 = 1 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 1+27+27 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-1)(27.5-27)(27.5-27) } ; ; T = sqrt{ 182.19 } = 13.5 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 13.5 }{ 1 } = 27 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 13.5 }{ 27 } = 1 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 13.5 }{ 27 } = 1 ; ;

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{ 1**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 2° 7'20" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-1**2-27**2 }{ 2 * 1 * 27 } ) = 88° 56'20" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-1**2-27**2 }{ 2 * 27 * 1 } ) = 88° 56'20" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 13.5 }{ 27.5 } = 0.49 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 1 }{ 2 * sin 2° 7'20" } = 13.5 ; ;




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

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