9 21 21 triangle

Acute isosceles triangle.

Sides: a = 9   b = 21   c = 21

Area: T = 92.30548617354
Perimeter: p = 51
Semiperimeter: s = 25.5

Angle ∠ A = α = 24.74772502324° = 24°44'50″ = 0.43219209974 rad
Angle ∠ B = β = 77.62663748838° = 77°37'35″ = 1.35548358281 rad
Angle ∠ C = γ = 77.62663748838° = 77°37'35″ = 1.35548358281 rad

Height: ha = 20.51221914968
Height: hb = 8.79109392129
Height: hc = 8.79109392129

Median: ma = 20.51221914968
Median: mb = 12.27880291578
Median: mc = 12.27880291578

Inradius: r = 3.62197984994
Circumradius: R = 10.75497046347

Vertex coordinates: A[21; 0] B[0; 0] C[1.92985714286; 8.79109392129]
Centroid: CG[7.64328571429; 2.9330313071]
Coordinates of the circumscribed circle: U[10.5; 2.3043508136]
Coordinates of the inscribed circle: I[4.5; 3.62197984994]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 155.2532749768° = 155°15'10″ = 0.43219209974 rad
∠ B' = β' = 102.3743625116° = 102°22'25″ = 1.35548358281 rad
∠ C' = γ' = 102.3743625116° = 102°22'25″ = 1.35548358281 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 = 9 ; ; b = 21 ; ; c = 21 ; ;

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

p = a+b+c = 9+21+21 = 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 * (25.5-9)(25.5-21)(25.5-21) } ; ; T = sqrt{ 8520.19 } = 92.3 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 92.3 }{ 9 } = 20.51 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 92.3 }{ 21 } = 8.79 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 92.3 }{ 21 } = 8.79 ; ;

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{ 9**2-21**2-21**2 }{ 2 * 21 * 21 } ) = 24° 44'50" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 21**2-9**2-21**2 }{ 2 * 9 * 21 } ) = 77° 37'35" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 21**2-9**2-21**2 }{ 2 * 21 * 9 } ) = 77° 37'35" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 92.3 }{ 25.5 } = 3.62 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 24° 44'50" } = 10.75 ; ;




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

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