16 18 25 triangle

Obtuse scalene triangle.

Sides: a = 16   b = 18   c = 25

Area: T = 143.5659874269
Perimeter: p = 59
Semiperimeter: s = 29.5

Angle ∠ A = α = 39.6466111147° = 39°38'46″ = 0.69219551751 rad
Angle ∠ B = β = 45.8733090067° = 45°52'23″ = 0.80106364597 rad
Angle ∠ C = γ = 94.4810798786° = 94°28'51″ = 1.64990010187 rad

Height: ha = 17.94549842836
Height: hb = 15.95110971409
Height: hc = 11.48547899415

Median: ma = 20.26107995894
Median: mb = 18.96604852259
Median: mc = 11.56550335062

Inradius: r = 4.86664364159
Circumradius: R = 12.53883224886

Vertex coordinates: A[25; 0] B[0; 0] C[11.14; 11.48547899415]
Centroid: CG[12.04766666667; 3.82882633138]
Coordinates of the circumscribed circle: U[12.5; -0.98795564444]
Coordinates of the inscribed circle: I[11.5; 4.86664364159]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140.3543888853° = 140°21'14″ = 0.69219551751 rad
∠ B' = β' = 134.1276909933° = 134°7'37″ = 0.80106364597 rad
∠ C' = γ' = 85.5199201214° = 85°31'9″ = 1.64990010187 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 = 18 ; ; c = 25 ; ;

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

p = a+b+c = 16+18+25 = 59 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 59 }{ 2 } = 29.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29.5 * (29.5-16)(29.5-18)(29.5-25) } ; ; T = sqrt{ 20609.44 } = 143.56 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 143.56 }{ 16 } = 17.94 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 143.56 }{ 18 } = 15.95 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 143.56 }{ 25 } = 11.48 ; ;

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**2-18**2-25**2 }{ 2 * 18 * 25 } ) = 39° 38'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-16**2-25**2 }{ 2 * 16 * 25 } ) = 45° 52'23" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-16**2-18**2 }{ 2 * 18 * 16 } ) = 94° 28'51" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 143.56 }{ 29.5 } = 4.87 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 16 }{ 2 * sin 39° 38'46" } = 12.54 ; ;




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

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