21 25 27 triangle

Acute scalene triangle.

Sides: a = 21   b = 25   c = 27

Area: T = 248.6132524825
Perimeter: p = 73
Semiperimeter: s = 36.5

Angle ∠ A = α = 47.44551038672° = 47°26'42″ = 0.82880732764 rad
Angle ∠ B = β = 61.27554267539° = 61°16'32″ = 1.06994579474 rad
Angle ∠ C = γ = 71.27994693789° = 71°16'46″ = 1.24440614297 rad

Height: ha = 23.67773833167
Height: hb = 19.8899001986
Height: hc = 18.41657425797

Median: ma = 23.80765117142
Median: mb = 20.70662792408
Median: mc = 18.72883208003

Inradius: r = 6.811130205
Circumradius: R = 14.25441088889

Vertex coordinates: A[27; 0] B[0; 0] C[10.09325925926; 18.41657425797]
Centroid: CG[12.36441975309; 6.13985808599]
Coordinates of the circumscribed circle: U[13.5; 4.57548901862]
Coordinates of the inscribed circle: I[11.5; 6.811130205]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 132.5554896133° = 132°33'18″ = 0.82880732764 rad
∠ B' = β' = 118.7254573246° = 118°43'28″ = 1.06994579474 rad
∠ C' = γ' = 108.7210530621° = 108°43'14″ = 1.24440614297 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 = 21 ; ; b = 25 ; ; c = 27 ; ;

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

p = a+b+c = 21+25+27 = 73 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 73 }{ 2 } = 36.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 36.5 * (36.5-21)(36.5-25)(36.5-27) } ; ; T = sqrt{ 61808.19 } = 248.61 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 248.61 }{ 21 } = 23.68 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 248.61 }{ 25 } = 19.89 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 248.61 }{ 27 } = 18.42 ; ;

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{ 21**2-25**2-27**2 }{ 2 * 25 * 27 } ) = 47° 26'42" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-21**2-27**2 }{ 2 * 21 * 27 } ) = 61° 16'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-21**2-25**2 }{ 2 * 25 * 21 } ) = 71° 16'46" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 248.61 }{ 36.5 } = 6.81 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 47° 26'42" } = 14.25 ; ;




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

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