Triangle calculator VC

Please enter the coordinates of the three vertices


Acute isosceles triangle.

Sides: a = 0.5   b = 0.44772135955   c = 0.5

Area: T = 0.1
Perimeter: p = 1.44772135955
Semiperimeter: s = 0.72436067977

Angle ∠ A = α = 63.43549488229° = 63°26'6″ = 1.10771487178 rad
Angle ∠ B = β = 53.13301023542° = 53°7'48″ = 0.9277295218 rad
Angle ∠ C = γ = 63.43549488229° = 63°26'6″ = 1.10771487178 rad

Height: ha = 0.4
Height: hb = 0.44772135955
Height: hc = 0.4

Median: ma = 0.40331128874
Median: mb = 0.44772135955
Median: mc = 0.40331128874

Inradius: r = 0.13881966011
Circumradius: R = 0.28795084972

Vertex coordinates: A[1.5; 0] B[2; 0] C[1.7; 0.4]
Centroid: CG[1.73333333333; 0.13333333333]
Coordinates of the circumscribed circle: U[0; 0]
Coordinates of the inscribed circle: I[0.10436474508; 0.13881966011]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 116.5655051177° = 116°33'54″ = 1.10771487178 rad
∠ B' = β' = 126.8769897646° = 126°52'12″ = 0.9277295218 rad
∠ C' = γ' = 116.5655051177° = 116°33'54″ = 1.10771487178 rad

Calculate another triangle




How did we calculate this triangle?

1. We compute side a from coordinates using the Pythagorean theorem

a = | beta gamma | = | beta - gamma | ; ; a**2 = ( beta _x- gamma _x)**2 + ( beta _y- gamma _y)**2 ; ; a = sqrt{ ( beta _x- gamma _x)**2 + ( beta _y- gamma _y)**2 } ; ; a = sqrt{ (2-1.7)**2 + (0-0.4)**2 } ; ; a = sqrt{ 0.25 } = 0.5 ; ;

2. We compute side b from coordinates using the Pythagorean theorem

b = | alpha gamma | = | alpha - gamma | ; ; b**2 = ( alpha _x- gamma _x)**2 + ( alpha _y- gamma _y)**2 ; ; b = sqrt{ ( alpha _x- gamma _x)**2 + ( alpha _y- gamma _y)**2 } ; ; b = sqrt{ (1.5-1.7)**2 + (0-0.4)**2 } ; ; b = sqrt{ 0.2 } = 0.45 ; ;

3. We compute side c from coordinates using the Pythagorean theorem

c = | alpha beta | = | alpha - beta | ; ; c**2 = ( alpha _x- beta _x)**2 + ( alpha _y- beta _y)**2 ; ; c = sqrt{ ( alpha _x- beta _x)**2 + ( alpha _y- beta _y)**2 } ; ; c = sqrt{ (1.5-2)**2 + (0-0)**2 } ; ; c = sqrt{ 0.25 } = 0.5 ; ;


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 = 0.5 ; ; b = 0.45 ; ; c = 0.5 ; ;

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

p = a+b+c = 0.5+0.45+0.5 = 1.45 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1.45 }{ 2 } = 0.72 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 0.72 * (0.72-0.5)(0.72-0.45)(0.72-0.5) } ; ; T = sqrt{ 0.01 } = 0.1 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 0.1 }{ 0.5 } = 0.4 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 0.1 }{ 0.45 } = 0.45 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 0.1 }{ 0.5 } = 0.4 ; ;

8. 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{ 0.5**2-0.45**2-0.5**2 }{ 2 * 0.45 * 0.5 } ) = 63° 26'6" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 0.45**2-0.5**2-0.5**2 }{ 2 * 0.5 * 0.5 } ) = 53° 7'48" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.5**2-0.5**2-0.45**2 }{ 2 * 0.45 * 0.5 } ) = 63° 26'6" ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 0.1 }{ 0.72 } = 0.14 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 0.5 }{ 2 * sin 63° 26'6" } = 0.28 ; ;




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

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