4 27 27 triangle

Acute isosceles triangle.

Sides: a = 4   b = 27   c = 27

Area: T = 53.85216480713
Perimeter: p = 58
Semiperimeter: s = 29

Angle ∠ A = α = 8.49660453335° = 8°29'46″ = 0.14882839645 rad
Angle ∠ B = β = 85.75219773333° = 85°45'7″ = 1.49766543446 rad
Angle ∠ C = γ = 85.75219773333° = 85°45'7″ = 1.49766543446 rad

Height: ha = 26.92658240357
Height: hb = 3.98990109682
Height: hc = 3.98990109682

Median: ma = 26.92658240357
Median: mb = 13.79331142241
Median: mc = 13.79331142241

Inradius: r = 1.85769533818
Circumradius: R = 13.53771901531

Vertex coordinates: A[27; 0] B[0; 0] C[0.29662962963; 3.98990109682]
Centroid: CG[9.09987654321; 1.33296703227]
Coordinates of the circumscribed circle: U[13.5; 1.00327548262]
Coordinates of the inscribed circle: I[2; 1.85769533818]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 171.5043954667° = 171°30'14″ = 0.14882839645 rad
∠ B' = β' = 94.24880226667° = 94°14'53″ = 1.49766543446 rad
∠ C' = γ' = 94.24880226667° = 94°14'53″ = 1.49766543446 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 = 4 ; ; b = 27 ; ; c = 27 ; ;

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

p = a+b+c = 4+27+27 = 58 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 58 }{ 2 } = 29 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 29 * (29-4)(29-27)(29-27) } ; ; T = sqrt{ 2900 } = 53.85 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 53.85 }{ 4 } = 26.93 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 53.85 }{ 27 } = 3.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 53.85 }{ 27 } = 3.99 ; ;

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{ 4**2-27**2-27**2 }{ 2 * 27 * 27 } ) = 8° 29'46" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 27**2-4**2-27**2 }{ 2 * 4 * 27 } ) = 85° 45'7" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 27**2-4**2-27**2 }{ 2 * 27 * 4 } ) = 85° 45'7" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 53.85 }{ 29 } = 1.86 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 4 }{ 2 * sin 8° 29'46" } = 13.54 ; ;




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

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