10 10 11 triangle

Acute isosceles triangle.

Sides: a = 10   b = 10   c = 11

Area: T = 45.93440559933
Perimeter: p = 31
Semiperimeter: s = 15.5

Angle ∠ A = α = 56.63329870308° = 56°37'59″ = 0.98884320889 rad
Angle ∠ B = β = 56.63329870308° = 56°37'59″ = 0.98884320889 rad
Angle ∠ C = γ = 66.73440259385° = 66°44'3″ = 1.16547284757 rad

Height: ha = 9.18768111987
Height: hb = 9.18768111987
Height: hc = 8.35216465442

Median: ma = 9.24766210045
Median: mb = 9.24766210045
Median: mc = 8.35216465442

Inradius: r = 2.96334874834
Circumradius: R = 5.98768434009

Vertex coordinates: A[11; 0] B[0; 0] C[5.5; 8.35216465442]
Centroid: CG[5.5; 2.78438821814]
Coordinates of the circumscribed circle: U[5.5; 2.36548031434]
Coordinates of the inscribed circle: I[5.5; 2.96334874834]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 123.3677012969° = 123°22'1″ = 0.98884320889 rad
∠ B' = β' = 123.3677012969° = 123°22'1″ = 0.98884320889 rad
∠ C' = γ' = 113.2665974062° = 113°15'57″ = 1.16547284757 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 = 10 ; ; b = 10 ; ; c = 11 ; ;

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

p = a+b+c = 10+10+11 = 31 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 31 }{ 2 } = 15.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 15.5 * (15.5-10)(15.5-10)(15.5-11) } ; ; T = sqrt{ 2109.94 } = 45.93 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 45.93 }{ 10 } = 9.19 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 45.93 }{ 10 } = 9.19 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 45.93 }{ 11 } = 8.35 ; ;

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{ 10**2-10**2-11**2 }{ 2 * 10 * 11 } ) = 56° 37'59" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 10**2-10**2-11**2 }{ 2 * 10 * 11 } ) = 56° 37'59" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 11**2-10**2-10**2 }{ 2 * 10 * 10 } ) = 66° 44'3" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 45.93 }{ 15.5 } = 2.96 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 56° 37'59" } = 5.99 ; ;




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

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