11 20 20 triangle

Acute isosceles triangle.

Sides: a = 11   b = 20   c = 20

Area: T = 105.7598864877
Perimeter: p = 51
Semiperimeter: s = 25.5

Angle ∠ A = α = 31.92440283257° = 31°55'27″ = 0.55771794048 rad
Angle ∠ B = β = 74.03879858372° = 74°2'17″ = 1.29222066244 rad
Angle ∠ C = γ = 74.03879858372° = 74°2'17″ = 1.29222066244 rad

Height: ha = 19.2298884523
Height: hb = 10.57658864877
Height: hc = 10.57658864877

Median: ma = 19.2298884523
Median: mb = 12.66988594593
Median: mc = 12.66988594593

Inradius: r = 4.14774064658
Circumradius: R = 10.40110193498

Vertex coordinates: A[20; 0] B[0; 0] C[3.025; 10.57658864877]
Centroid: CG[7.675; 3.52552954959]
Coordinates of the circumscribed circle: U[10; 2.86602803212]
Coordinates of the inscribed circle: I[5.5; 4.14774064658]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 148.0765971674° = 148°4'33″ = 0.55771794048 rad
∠ B' = β' = 105.9622014163° = 105°57'43″ = 1.29222066244 rad
∠ C' = γ' = 105.9622014163° = 105°57'43″ = 1.29222066244 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 = 11 ; ; b = 20 ; ; c = 20 ; ;

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

p = a+b+c = 11+20+20 = 51 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 51 }{ 2 } = 25.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 25.5 * (25.5-11)(25.5-20)(25.5-20) } ; ; T = sqrt{ 11184.94 } = 105.76 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 105.76 }{ 11 } = 19.23 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 105.76 }{ 20 } = 10.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 105.76 }{ 20 } = 10.58 ; ;

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{ 11**2-20**2-20**2 }{ 2 * 20 * 20 } ) = 31° 55'27" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-11**2-20**2 }{ 2 * 11 * 20 } ) = 74° 2'17" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-11**2-20**2 }{ 2 * 20 * 11 } ) = 74° 2'17" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 105.76 }{ 25.5 } = 4.15 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 11 }{ 2 * sin 31° 55'27" } = 10.4 ; ;




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

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