Triangle calculator VC

Please enter the coordinates of the three vertices


Acute isosceles triangle.

Sides: a = 11.4021754251   b = 11.4021754251   c = 2.82884271247

Area: T = 16
Perimeter: p = 25.63219356267
Semiperimeter: s = 12.81659678134

Angle ∠ A = α = 82.87549836511° = 82°52'30″ = 1.44664413322 rad
Angle ∠ B = β = 82.87549836511° = 82°52'30″ = 1.44664413322 rad
Angle ∠ C = γ = 14.25500326978° = 14°15' = 0.24987099891 rad

Height: ha = 2.80765856618
Height: hb = 2.80765856618
Height: hc = 11.3143708499

Median: ma = 6.04215229868
Median: mb = 6.04215229868
Median: mc = 11.3143708499

Inradius: r = 1.24884425861
Circumradius: R = 5.74552425971

Vertex coordinates: A[2; 4] B[4; 2] C[-5; -5]
Centroid: CG[0.33333333333; 0.33333333333]
Coordinates of the circumscribed circle: U[0; 0]
Coordinates of the inscribed circle: I[0.15660553233; 1.24884425861]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 97.12550163489° = 97°7'30″ = 1.44664413322 rad
∠ B' = β' = 97.12550163489° = 97°7'30″ = 1.44664413322 rad
∠ C' = γ' = 165.7549967302° = 165°45' = 0.24987099891 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{ (4-(-5))**2 + (2-(-5))**2 } ; ; a = sqrt{ 130 } = 11.4 ; ;

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{ (2-(-5))**2 + (4-(-5))**2 } ; ; b = sqrt{ 130 } = 11.4 ; ;

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


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 = 11.4 ; ; b = 11.4 ; ; c = 2.83 ; ;

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

p = a+b+c = 11.4+11.4+2.83 = 25.63 ; ;

5. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 25.63 }{ 2 } = 12.82 ; ;

6. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 12.82 * (12.82-11.4)(12.82-11.4)(12.82-2.83) } ; ; T = sqrt{ 256 } = 16 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 16 }{ 11.4 } = 2.81 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 16 }{ 11.4 } = 2.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 16 }{ 2.83 } = 11.31 ; ;

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{ 11.4**2-11.4**2-2.83**2 }{ 2 * 11.4 * 2.83 } ) = 82° 52'30" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 11.4**2-11.4**2-2.83**2 }{ 2 * 11.4 * 2.83 } ) = 82° 52'30" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 2.83**2-11.4**2-11.4**2 }{ 2 * 11.4 * 11.4 } ) = 14° 15' ; ;

9. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 16 }{ 12.82 } = 1.25 ; ;

10. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11.4 }{ 2 * sin 82° 52'30" } = 5.75 ; ;




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

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