5 10 10 triangle

Acute isosceles triangle.

Sides: a = 5   b = 10   c = 10

Area: T = 24.20661459138
Perimeter: p = 25
Semiperimeter: s = 12.5

Angle ∠ A = α = 28.95550243719° = 28°57'18″ = 0.50553605103 rad
Angle ∠ B = β = 75.52224878141° = 75°31'21″ = 1.31881160717 rad
Angle ∠ C = γ = 75.52224878141° = 75°31'21″ = 1.31881160717 rad

Height: ha = 9.68224583655
Height: hb = 4.84112291828
Height: hc = 4.84112291828

Median: ma = 9.68224583655
Median: mb = 6.1243724357
Median: mc = 6.1243724357

Inradius: r = 1.93664916731
Circumradius: R = 5.16439777949

Vertex coordinates: A[10; 0] B[0; 0] C[1.25; 4.84112291828]
Centroid: CG[3.75; 1.61437430609]
Coordinates of the circumscribed circle: U[5; 1.29109944487]
Coordinates of the inscribed circle: I[2.5; 1.93664916731]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 151.0454975628° = 151°2'42″ = 0.50553605103 rad
∠ B' = β' = 104.4787512186° = 104°28'39″ = 1.31881160717 rad
∠ C' = γ' = 104.4787512186° = 104°28'39″ = 1.31881160717 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 = 5 ; ; b = 10 ; ; c = 10 ; ;

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

p = a+b+c = 5+10+10 = 25 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 25 }{ 2 } = 12.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 12.5 * (12.5-5)(12.5-10)(12.5-10) } ; ; T = sqrt{ 585.94 } = 24.21 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 24.21 }{ 5 } = 9.68 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 24.21 }{ 10 } = 4.84 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 24.21 }{ 10 } = 4.84 ; ;

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{ 5**2-10**2-10**2 }{ 2 * 10 * 10 } ) = 28° 57'18" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 10**2-5**2-10**2 }{ 2 * 5 * 10 } ) = 75° 31'21" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 10**2-5**2-10**2 }{ 2 * 10 * 5 } ) = 75° 31'21" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 24.21 }{ 12.5 } = 1.94 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 5 }{ 2 * sin 28° 57'18" } = 5.16 ; ;




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

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