9 23 25 triangle

Obtuse scalene triangle.

Sides: a = 9   b = 23   c = 25

Area: T = 103.4322042907
Perimeter: p = 57
Semiperimeter: s = 28.5

Angle ∠ A = α = 21.08656802932° = 21°5'8″ = 0.36880145461 rad
Angle ∠ B = β = 66.83879312011° = 66°50'17″ = 1.16765419647 rad
Angle ∠ C = γ = 92.07663885057° = 92°4'35″ = 1.60770361428 rad

Height: ha = 22.98548984239
Height: hb = 8.99440906876
Height: hc = 8.27545634326

Median: ma = 23.5965550428
Median: mb = 14.85876579581
Median: mc = 12.19663109177

Inradius: r = 3.6299194488
Circumradius: R = 12.50882127708

Vertex coordinates: A[25; 0] B[0; 0] C[3.54; 8.27545634326]
Centroid: CG[9.51333333333; 2.75881878109]
Coordinates of the circumscribed circle: U[12.5; -0.45331961149]
Coordinates of the inscribed circle: I[5.5; 3.6299194488]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 158.9144319707° = 158°54'52″ = 0.36880145461 rad
∠ B' = β' = 113.1622068799° = 113°9'43″ = 1.16765419647 rad
∠ C' = γ' = 87.92436114943° = 87°55'25″ = 1.60770361428 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 = 23 ; ; c = 25 ; ;

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

p = a+b+c = 9+23+25 = 57 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 57 }{ 2 } = 28.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 28.5 * (28.5-9)(28.5-23)(28.5-25) } ; ; T = sqrt{ 10698.19 } = 103.43 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 103.43 }{ 9 } = 22.98 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 103.43 }{ 23 } = 8.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 103.43 }{ 25 } = 8.27 ; ;

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**2-23**2-25**2 }{ 2 * 23 * 25 } ) = 21° 5'8" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 23**2-9**2-25**2 }{ 2 * 9 * 25 } ) = 66° 50'17" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-9**2-23**2 }{ 2 * 23 * 9 } ) = 92° 4'35" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 103.43 }{ 28.5 } = 3.63 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 9 }{ 2 * sin 21° 5'8" } = 12.51 ; ;




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

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