12 25 25 triangle

Acute isosceles triangle.

Sides: a = 12   b = 25   c = 25

Area: T = 145.6165933194
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 27.77330807253° = 27°46'23″ = 0.48547317021 rad
Angle ∠ B = β = 76.11334596374° = 76°6'48″ = 1.32884304758 rad
Angle ∠ C = γ = 76.11334596374° = 76°6'48″ = 1.32884304758 rad

Height: ha = 24.2699322199
Height: hb = 11.64992746555
Height: hc = 11.64992746555

Median: ma = 24.2699322199
Median: mb = 15.10879449297
Median: mc = 15.10879449297

Inradius: r = 4.69772881676
Circumradius: R = 12.87663381786

Vertex coordinates: A[25; 0] B[0; 0] C[2.88; 11.64992746555]
Centroid: CG[9.29333333333; 3.88330915518]
Coordinates of the circumscribed circle: U[12.5; 3.09903211629]
Coordinates of the inscribed circle: I[6; 4.69772881676]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 152.2276919275° = 152°13'37″ = 0.48547317021 rad
∠ B' = β' = 103.8876540363° = 103°53'12″ = 1.32884304758 rad
∠ C' = γ' = 103.8876540363° = 103°53'12″ = 1.32884304758 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 = 12 ; ; b = 25 ; ; c = 25 ; ;

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

p = a+b+c = 12+25+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 * (31-12)(31-25)(31-25) } ; ; T = sqrt{ 21204 } = 145.62 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 145.62 }{ 12 } = 24.27 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 145.62 }{ 25 } = 11.65 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 145.62 }{ 25 } = 11.65 ; ;

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{ 12**2-25**2-25**2 }{ 2 * 25 * 25 } ) = 27° 46'23" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-12**2-25**2 }{ 2 * 12 * 25 } ) = 76° 6'48" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-12**2-25**2 }{ 2 * 25 * 12 } ) = 76° 6'48" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 145.62 }{ 31 } = 4.7 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 12 }{ 2 * sin 27° 46'23" } = 12.88 ; ;




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

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