24 25 25 triangle

Acute isosceles triangle.

Sides: a = 24   b = 25   c = 25

Area: T = 263.1810546393
Perimeter: p = 74
Semiperimeter: s = 37

Angle ∠ A = α = 57.37108040282° = 57°22'15″ = 1.00113094248 rad
Angle ∠ B = β = 61.31545979859° = 61°18'53″ = 1.07701416144 rad
Angle ∠ C = γ = 61.31545979859° = 61°18'53″ = 1.07701416144 rad

Height: ha = 21.93217121995
Height: hb = 21.05444437115
Height: hc = 21.05444437115

Median: ma = 21.93217121995
Median: mb = 21.07772389084
Median: mc = 21.07772389084

Inradius: r = 7.11329877404
Circumradius: R = 14.24987735184

Vertex coordinates: A[25; 0] B[0; 0] C[11.52; 21.05444437115]
Centroid: CG[12.17333333333; 7.01881479038]
Coordinates of the circumscribed circle: U[12.5; 6.83994112888]
Coordinates of the inscribed circle: I[12; 7.11329877404]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 122.6299195972° = 122°37'45″ = 1.00113094248 rad
∠ B' = β' = 118.6855402014° = 118°41'7″ = 1.07701416144 rad
∠ C' = γ' = 118.6855402014° = 118°41'7″ = 1.07701416144 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 = 24 ; ; b = 25 ; ; c = 25 ; ;

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

p = a+b+c = 24+25+25 = 74 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 74 }{ 2 } = 37 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 37 * (37-24)(37-25)(37-25) } ; ; T = sqrt{ 69264 } = 263.18 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 263.18 }{ 24 } = 21.93 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 263.18 }{ 25 } = 21.05 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 263.18 }{ 25 } = 21.05 ; ;

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{ 24**2-25**2-25**2 }{ 2 * 25 * 25 } ) = 57° 22'15" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-24**2-25**2 }{ 2 * 24 * 25 } ) = 61° 18'53" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-24**2-25**2 }{ 2 * 25 * 24 } ) = 61° 18'53" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 263.18 }{ 37 } = 7.11 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 24 }{ 2 * sin 57° 22'15" } = 14.25 ; ;




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

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