12 20 20 triangle

Acute isosceles triangle.

Sides: a = 12   b = 20   c = 20

Area: T = 114.473270417
Perimeter: p = 52
Semiperimeter: s = 26

Angle ∠ A = α = 34.91552062474° = 34°54'55″ = 0.6099385308 rad
Angle ∠ B = β = 72.54223968763° = 72°32'33″ = 1.26661036728 rad
Angle ∠ C = γ = 72.54223968763° = 72°32'33″ = 1.26661036728 rad

Height: ha = 19.07987840283
Height: hb = 11.4477270417
Height: hc = 11.4477270417

Median: ma = 19.07987840283
Median: mb = 13.11548770486
Median: mc = 13.11548770486

Inradius: r = 4.40327963142
Circumradius: R = 10.48328483672

Vertex coordinates: A[20; 0] B[0; 0] C[3.6; 11.4477270417]
Centroid: CG[7.86766666667; 3.81657568057]
Coordinates of the circumscribed circle: U[10; 3.14548545102]
Coordinates of the inscribed circle: I[6; 4.40327963142]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 145.0854793753° = 145°5'5″ = 0.6099385308 rad
∠ B' = β' = 107.4587603124° = 107°27'27″ = 1.26661036728 rad
∠ C' = γ' = 107.4587603124° = 107°27'27″ = 1.26661036728 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 = 12 ; ; b = 20 ; ; c = 20 ; ;

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

p = a+b+c = 12+20+20 = 52 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 52 }{ 2 } = 26 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 26 * (26-12)(26-20)(26-20) } ; ; T = sqrt{ 13104 } = 114.47 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 114.47 }{ 12 } = 19.08 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 114.47 }{ 20 } = 11.45 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 114.47 }{ 20 } = 11.45 ; ;

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{ 12**2-20**2-20**2 }{ 2 * 20 * 20 } ) = 34° 54'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-12**2-20**2 }{ 2 * 12 * 20 } ) = 72° 32'33" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-12**2-20**2 }{ 2 * 20 * 12 } ) = 72° 32'33" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 114.47 }{ 26 } = 4.4 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 12 }{ 2 * sin 34° 54'55" } = 10.48 ; ;




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

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