0.707 0.707 0.5 triangle

Acute isosceles triangle.

Sides: a = 0.707   b = 0.707   c = 0.5

Area: T = 0.16553309182
Perimeter: p = 1.914
Semiperimeter: s = 0.957

Angle ∠ A = α = 69.29219181481° = 69°17'31″ = 1.20993721167 rad
Angle ∠ B = β = 69.29219181481° = 69°17'31″ = 1.20993721167 rad
Angle ∠ C = γ = 41.41661637038° = 41°24'58″ = 0.72328484202 rad

Height: ha = 0.46876970811
Height: hb = 0.46876970811
Height: hc = 0.66113236726

Median: ma = 0.54999622486
Median: mb = 0.54999622486
Median: mc = 0.66113236726

Inradius: r = 0.17327595801
Circumradius: R = 0.37879155508

Vertex coordinates: A[0.5; 0] B[0; 0] C[0.25; 0.66113236726]
Centroid: CG[0.25; 0.22204412242]
Coordinates of the circumscribed circle: U[0.25; 0.28334081219]
Coordinates of the inscribed circle: I[0.25; 0.17327595801]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 110.7088081852° = 110°42'29″ = 1.20993721167 rad
∠ B' = β' = 110.7088081852° = 110°42'29″ = 1.20993721167 rad
∠ C' = γ' = 138.5843836296° = 138°35'2″ = 0.72328484202 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 = 0.71 ; ; b = 0.71 ; ; c = 0.5 ; ;

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

p = a+b+c = 0.71+0.71+0.5 = 1.91 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1.91 }{ 2 } = 0.96 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 0.96 * (0.96-0.71)(0.96-0.71)(0.96-0.5) } ; ; T = sqrt{ 0.03 } = 0.17 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 0.17 }{ 0.71 } = 0.47 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 0.17 }{ 0.71 } = 0.47 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 0.17 }{ 0.5 } = 0.66 ; ;

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{ 0.71**2-0.71**2-0.5**2 }{ 2 * 0.71 * 0.5 } ) = 69° 17'31" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 0.71**2-0.71**2-0.5**2 }{ 2 * 0.71 * 0.5 } ) = 69° 17'31" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 0.5**2-0.71**2-0.71**2 }{ 2 * 0.71 * 0.71 } ) = 41° 24'58" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 0.17 }{ 0.96 } = 0.17 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 0.71 }{ 2 * sin 69° 17'31" } = 0.38 ; ;




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

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