16 21 21 triangle

Acute isosceles triangle.

Sides: a = 16   b = 21   c = 21

Area: T = 155.3321902712
Perimeter: p = 58
Semiperimeter: s = 29

Angle ∠ A = α = 44.78553756108° = 44°47'7″ = 0.78216522612 rad
Angle ∠ B = β = 67.60773121946° = 67°36'26″ = 1.18799701962 rad
Angle ∠ C = γ = 67.60773121946° = 67°36'26″ = 1.18799701962 rad

Height: ha = 19.41664878389
Height: hb = 14.7943514544
Height: hc = 14.7943514544

Median: ma = 19.41664878389
Median: mb = 15.43553490404
Median: mc = 15.43553490404

Inradius: r = 5.35662725073
Circumradius: R = 11.35663277679

Vertex coordinates: A[21; 0] B[0; 0] C[6.09552380952; 14.7943514544]
Centroid: CG[9.03217460317; 4.93111715147]
Coordinates of the circumscribed circle: U[10.5; 4.3266220102]
Coordinates of the inscribed circle: I[8; 5.35662725073]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135.2154624389° = 135°12'53″ = 0.78216522612 rad
∠ B' = β' = 112.3932687805° = 112°23'34″ = 1.18799701962 rad
∠ C' = γ' = 112.3932687805° = 112°23'34″ = 1.18799701962 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 = 16 ; ; b = 21 ; ; c = 21 ; ;

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

p = a+b+c = 16+21+21 = 58 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 58 }{ 2 } = 29 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29 * (29-16)(29-21)(29-21) } ; ; T = sqrt{ 24128 } = 155.33 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 155.33 }{ 16 } = 19.42 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 155.33 }{ 21 } = 14.79 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 155.33 }{ 21 } = 14.79 ; ;

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{ 16**2-21**2-21**2 }{ 2 * 21 * 21 } ) = 44° 47'7" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-16**2-21**2 }{ 2 * 16 * 21 } ) = 67° 36'26" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-16**2-21**2 }{ 2 * 21 * 16 } ) = 67° 36'26" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 155.33 }{ 29 } = 5.36 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 44° 47'7" } = 11.36 ; ;




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

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