Triangle calculator VC

Please enter the coordinates of the three vertices


Acute isosceles triangle.

Sides: a = 2.82884271247   b = 3.16222776602   c = 3.16222776602

Area: T = 4
Perimeter: p = 9.15329824451
Semiperimeter: s = 4.57664912225

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

Height: ha = 2.82884271247
Height: hb = 2.53298221281
Height: hc = 2.53298221281

Median: ma = 2.82884271247
Median: mb = 2.55495097568
Median: mc = 2.55495097568

Inradius: r = 0.87440320489
Circumradius: R = 1.7687766953

Vertex coordinates: A[1; 1] B[2; 4] C[4; 2]
Centroid: CG[2.33333333333; 2.33333333333]
Coordinates of the circumscribed circle: U[0; 0]
Coordinates of the inscribed circle: I[0.43770160244; 0.87440320489]

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

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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{ (2-4)**2 + (4-2)**2 } ; ; a = sqrt{ 8 } = 2.83 ; ;

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{ (1-4)**2 + (1-2)**2 } ; ; b = sqrt{ 10 } = 3.16 ; ;

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{ (1-2)**2 + (1-4)**2 } ; ; c = sqrt{ 10 } = 3.16 ; ;


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 = 2.83 ; ; b = 3.16 ; ; c = 3.16 ; ;

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

p = a+b+c = 2.83+3.16+3.16 = 9.15 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 9.15 }{ 2 } = 4.58 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 4.58 * (4.58-2.83)(4.58-3.16)(4.58-3.16) } ; ; T = sqrt{ 16 } = 4 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 4 }{ 2.83 } = 2.83 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 4 }{ 3.16 } = 2.53 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 4 }{ 3.16 } = 2.53 ; ;

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{ 2.83**2-3.16**2-3.16**2 }{ 2 * 3.16 * 3.16 } ) = 53° 7'48" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 3.16**2-2.83**2-3.16**2 }{ 2 * 2.83 * 3.16 } ) = 63° 26'6" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 3.16**2-2.83**2-3.16**2 }{ 2 * 3.16 * 2.83 } ) = 63° 26'6" ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 4 }{ 4.58 } = 0.87 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 2.83 }{ 2 * sin 53° 7'48" } = 1.77 ; ;




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