10 22 29 triangle

Obtuse scalene triangle.

Sides: a = 10   b = 22   c = 29

Area: T = 89.28657071429
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 16.25437888126° = 16°15'14″ = 0.28436821307 rad
Angle ∠ B = β = 38.00774183478° = 38°27″ = 0.66333545904 rad
Angle ∠ C = γ = 125.739879284° = 125°44'20″ = 2.19545559325 rad

Height: ha = 17.85771414286
Height: hb = 8.11768824675
Height: hc = 6.15876349754

Median: ma = 25.24987623459
Median: mb = 18.69549190958
Median: mc = 9.042157066

Inradius: r = 2.92774002342
Circumradius: R = 17.86440014291

Vertex coordinates: A[29; 0] B[0; 0] C[7.87993103448; 6.15876349754]
Centroid: CG[12.29331034483; 2.05325449918]
Coordinates of the circumscribed circle: U[14.5; -10.43442008347]
Coordinates of the inscribed circle: I[8.5; 2.92774002342]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 163.7466211187° = 163°44'46″ = 0.28436821307 rad
∠ B' = β' = 141.9932581652° = 141°59'33″ = 0.66333545904 rad
∠ C' = γ' = 54.26112071604° = 54°15'40″ = 2.19545559325 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 = 10 ; ; b = 22 ; ; c = 29 ; ;

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

p = a+b+c = 10+22+29 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-10)(30.5-22)(30.5-29) } ; ; T = sqrt{ 7971.94 } = 89.29 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 89.29 }{ 10 } = 17.86 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 89.29 }{ 22 } = 8.12 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 89.29 }{ 29 } = 6.16 ; ;

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{ 10**2-22**2-29**2 }{ 2 * 22 * 29 } ) = 16° 15'14" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 22**2-10**2-29**2 }{ 2 * 10 * 29 } ) = 38° 27" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-10**2-22**2 }{ 2 * 22 * 10 } ) = 125° 44'20" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 89.29 }{ 30.5 } = 2.93 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 16° 15'14" } = 17.86 ; ;




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

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