27 29 29 triangle

Acute isosceles triangle.

Sides: a = 27   b = 29   c = 29

Area: T = 346.4932694728
Perimeter: p = 85
Semiperimeter: s = 42.5

Angle ∠ A = α = 55.4877404317° = 55°29'15″ = 0.96884378987 rad
Angle ∠ B = β = 62.25662978415° = 62°15'23″ = 1.08765773774 rad
Angle ∠ C = γ = 62.25662978415° = 62°15'23″ = 1.08765773774 rad

Height: ha = 25.66661255354
Height: hb = 23.89660479123
Height: hc = 23.89660479123

Median: ma = 25.66661255354
Median: mb = 23.97439441895
Median: mc = 23.97439441895

Inradius: r = 8.15327692877
Circumradius: R = 16.38334622962

Vertex coordinates: A[29; 0] B[0; 0] C[12.56989655172; 23.89660479123]
Centroid: CG[13.85663218391; 7.96553493041]
Coordinates of the circumscribed circle: U[14.5; 7.62767841724]
Coordinates of the inscribed circle: I[13.5; 8.15327692877]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 124.5132595683° = 124°30'45″ = 0.96884378987 rad
∠ B' = β' = 117.7443702158° = 117°44'37″ = 1.08765773774 rad
∠ C' = γ' = 117.7443702158° = 117°44'37″ = 1.08765773774 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 = 27 ; ; b = 29 ; ; c = 29 ; ;

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

p = a+b+c = 27+29+29 = 85 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 85 }{ 2 } = 42.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 42.5 * (42.5-27)(42.5-29)(42.5-29) } ; ; T = sqrt{ 120057.19 } = 346.49 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 346.49 }{ 27 } = 25.67 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 346.49 }{ 29 } = 23.9 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 346.49 }{ 29 } = 23.9 ; ;

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{ 27**2-29**2-29**2 }{ 2 * 29 * 29 } ) = 55° 29'15" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 29**2-27**2-29**2 }{ 2 * 27 * 29 } ) = 62° 15'23" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 29**2-27**2-29**2 }{ 2 * 29 * 27 } ) = 62° 15'23" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 346.49 }{ 42.5 } = 8.15 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 27 }{ 2 * sin 55° 29'15" } = 16.38 ; ;




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

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