6 25 30 triangle

Obtuse scalene triangle.

Sides: a = 6   b = 25   c = 30

Area: T = 45.33114184645
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 6.94331109138° = 6°56'35″ = 0.12111801458 rad
Angle ∠ B = β = 30.2443927956° = 30°14'38″ = 0.52878561216 rad
Angle ∠ C = γ = 142.813296113° = 142°48'47″ = 2.49325563862 rad

Height: ha = 15.11104728215
Height: hb = 3.62765134772
Height: hc = 3.02220945643

Median: ma = 27.45499544626
Median: mb = 17.6566443583
Median: mc = 10.27113192921

Inradius: r = 1.48662760152
Circumradius: R = 24.81772247441

Vertex coordinates: A[30; 0] B[0; 0] C[5.18333333333; 3.02220945643]
Centroid: CG[11.72877777778; 1.00773648548]
Coordinates of the circumscribed circle: U[15; -19.77110557128]
Coordinates of the inscribed circle: I[5.5; 1.48662760152]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 173.0576889086° = 173°3'25″ = 0.12111801458 rad
∠ B' = β' = 149.7566072044° = 149°45'22″ = 0.52878561216 rad
∠ C' = γ' = 37.18770388698° = 37°11'13″ = 2.49325563862 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 = 6 ; ; b = 25 ; ; c = 30 ; ;

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

p = a+b+c = 6+25+30 = 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 * (30.5-6)(30.5-25)(30.5-30) } ; ; T = sqrt{ 2054.94 } = 45.33 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 45.33 }{ 6 } = 15.11 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 45.33 }{ 25 } = 3.63 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 45.33 }{ 30 } = 3.02 ; ;

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{ 6**2-25**2-30**2 }{ 2 * 25 * 30 } ) = 6° 56'35" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-6**2-30**2 }{ 2 * 6 * 30 } ) = 30° 14'38" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 30**2-6**2-25**2 }{ 2 * 25 * 6 } ) = 142° 48'47" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 45.33 }{ 30.5 } = 1.49 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 6 }{ 2 * sin 6° 56'35" } = 24.82 ; ;




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

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