3 22 22 triangle

Acute isosceles triangle.

Sides: a = 3   b = 22   c = 22

Area: T = 32.92332061015
Perimeter: p = 47
Semiperimeter: s = 23.5

Angle ∠ A = α = 7.81991270493° = 7°49'9″ = 0.13664695116 rad
Angle ∠ B = β = 86.09904364753° = 86°5'26″ = 1.5032561571 rad
Angle ∠ C = γ = 86.09904364753° = 86°5'26″ = 1.5032561571 rad

Height: ha = 21.94988040676
Height: hb = 2.99330187365
Height: hc = 2.99330187365

Median: ma = 21.94988040676
Median: mb = 11.20326782512
Median: mc = 11.20326782512

Inradius: r = 1.40109874937
Circumradius: R = 11.02656576738

Vertex coordinates: A[22; 0] B[0; 0] C[0.20545454545; 2.99330187365]
Centroid: CG[7.40215151515; 0.99876729122]
Coordinates of the circumscribed circle: U[11; 0.75217493869]
Coordinates of the inscribed circle: I[1.5; 1.40109874937]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 172.1810872951° = 172°10'51″ = 0.13664695116 rad
∠ B' = β' = 93.91095635247° = 93°54'34″ = 1.5032561571 rad
∠ C' = γ' = 93.91095635247° = 93°54'34″ = 1.5032561571 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 = 3 ; ; b = 22 ; ; c = 22 ; ;

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

p = a+b+c = 3+22+22 = 47 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 47 }{ 2 } = 23.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 23.5 * (23.5-3)(23.5-22)(23.5-22) } ; ; T = sqrt{ 1083.94 } = 32.92 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 32.92 }{ 3 } = 21.95 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 32.92 }{ 22 } = 2.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 32.92 }{ 22 } = 2.99 ; ;

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{ 3**2-22**2-22**2 }{ 2 * 22 * 22 } ) = 7° 49'9" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-3**2-22**2 }{ 2 * 3 * 22 } ) = 86° 5'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 22**2-3**2-22**2 }{ 2 * 22 * 3 } ) = 86° 5'26" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 32.92 }{ 23.5 } = 1.4 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 3 }{ 2 * sin 7° 49'9" } = 11.03 ; ;




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

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