20 27 27 triangle

Acute isosceles triangle.

Sides: a = 20   b = 27   c = 27

Area: T = 250.799872408
Perimeter: p = 74
Semiperimeter: s = 37

Angle ∠ A = α = 43.4776921583° = 43°28'37″ = 0.75988154303 rad
Angle ∠ B = β = 68.26215392085° = 68°15'42″ = 1.19113886117 rad
Angle ∠ C = γ = 68.26215392085° = 68°15'42″ = 1.19113886117 rad

Height: ha = 25.0879872408
Height: hb = 18.57876832652
Height: hc = 18.57876832652

Median: ma = 25.0879872408
Median: mb = 19.5511214796
Median: mc = 19.5511214796

Inradius: r = 6.7788343894
Circumradius: R = 14.53435667611

Vertex coordinates: A[27; 0] B[0; 0] C[7.40774074074; 18.57876832652]
Centroid: CG[11.46991358025; 6.19325610884]
Coordinates of the circumscribed circle: U[13.5; 5.38328025041]
Coordinates of the inscribed circle: I[10; 6.7788343894]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 136.5233078417° = 136°31'23″ = 0.75988154303 rad
∠ B' = β' = 111.7388460792° = 111°44'18″ = 1.19113886117 rad
∠ C' = γ' = 111.7388460792° = 111°44'18″ = 1.19113886117 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 = 20 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 20+27+27 = 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 * (37-20)(37-27)(37-27) } ; ; T = sqrt{ 62900 } = 250.8 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 250.8 }{ 20 } = 25.08 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 250.8 }{ 27 } = 18.58 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 250.8 }{ 27 } = 18.58 ; ;

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{ 20**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 43° 28'37" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-20**2-27**2 }{ 2 * 20 * 27 } ) = 68° 15'42" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-20**2-27**2 }{ 2 * 27 * 20 } ) = 68° 15'42" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 250.8 }{ 37 } = 6.78 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 20 }{ 2 * sin 43° 28'37" } = 14.53 ; ;




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

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