17 20 25 triangle

Acute scalene triangle.

Sides: a = 17   b = 20   c = 25

Area: T = 169.2455383984
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 42.60882175298° = 42°36'30″ = 0.74436536843 rad
Angle ∠ B = β = 52.79223196383° = 52°47'32″ = 0.92113997975 rad
Angle ∠ C = γ = 84.59994628319° = 84°35'58″ = 1.47765391718 rad

Height: ha = 19.91112216452
Height: hb = 16.92545383984
Height: hc = 13.54396307187

Median: ma = 20.98221352584
Median: mb = 18.89444436277
Median: mc = 13.7220422734

Inradius: r = 5.46595285156
Circumradius: R = 12.55657338698

Vertex coordinates: A[25; 0] B[0; 0] C[10.28; 13.54396307187]
Centroid: CG[11.76; 4.51332102396]
Coordinates of the circumscribed circle: U[12.5; 1.18217161289]
Coordinates of the inscribed circle: I[11; 5.46595285156]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 137.392178247° = 137°23'30″ = 0.74436536843 rad
∠ B' = β' = 127.2087680362° = 127°12'28″ = 0.92113997975 rad
∠ C' = γ' = 95.40105371681° = 95°24'2″ = 1.47765391718 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 = 17 ; ; b = 20 ; ; c = 25 ; ;

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

p = a+b+c = 17+20+25 = 62 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 62 }{ 2 } = 31 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31 * (31-17)(31-20)(31-25) } ; ; T = sqrt{ 28644 } = 169.25 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 169.25 }{ 17 } = 19.91 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 169.25 }{ 20 } = 16.92 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 169.25 }{ 25 } = 13.54 ; ;

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{ 17**2-20**2-25**2 }{ 2 * 20 * 25 } ) = 42° 36'30" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-17**2-25**2 }{ 2 * 17 * 25 } ) = 52° 47'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-17**2-20**2 }{ 2 * 20 * 17 } ) = 84° 35'58" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 169.25 }{ 31 } = 5.46 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 17 }{ 2 * sin 42° 36'30" } = 12.56 ; ;




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

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