15 20 20 triangle

Acute isosceles triangle.

Sides: a = 15   b = 20   c = 20

Area: T = 139.0543721633
Perimeter: p = 55
Semiperimeter: s = 27.5

Angle ∠ A = α = 44.04986256741° = 44°2'55″ = 0.7698793549 rad
Angle ∠ B = β = 67.9765687163° = 67°58'32″ = 1.18663995523 rad
Angle ∠ C = γ = 67.9765687163° = 67°58'32″ = 1.18663995523 rad

Height: ha = 18.54404962177
Height: hb = 13.90553721633
Height: hc = 13.90553721633

Median: ma = 18.54404962177
Median: mb = 14.57773797371
Median: mc = 14.57773797371

Inradius: r = 5.05664989685
Circumradius: R = 10.78771977994

Vertex coordinates: A[20; 0] B[0; 0] C[5.625; 13.90553721633]
Centroid: CG[8.54216666667; 4.63551240544]
Coordinates of the circumscribed circle: U[10; 4.04551991748]
Coordinates of the inscribed circle: I[7.5; 5.05664989685]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 135.9511374326° = 135°57'5″ = 0.7698793549 rad
∠ B' = β' = 112.0244312837° = 112°1'28″ = 1.18663995523 rad
∠ C' = γ' = 112.0244312837° = 112°1'28″ = 1.18663995523 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 = 15 ; ; b = 20 ; ; c = 20 ; ;

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

p = a+b+c = 15+20+20 = 55 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 55 }{ 2 } = 27.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 27.5 * (27.5-15)(27.5-20)(27.5-20) } ; ; T = sqrt{ 19335.94 } = 139.05 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 139.05 }{ 15 } = 18.54 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 139.05 }{ 20 } = 13.91 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 139.05 }{ 20 } = 13.91 ; ;

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{ 15**2-20**2-20**2 }{ 2 * 20 * 20 } ) = 44° 2'55" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 20**2-15**2-20**2 }{ 2 * 15 * 20 } ) = 67° 58'32" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 20**2-15**2-20**2 }{ 2 * 20 * 15 } ) = 67° 58'32" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 139.05 }{ 27.5 } = 5.06 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 44° 2'55" } = 10.79 ; ;




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

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