300 300 424.26 triangle

Acute isosceles triangle.

Sides: a = 300   b = 300   c = 424.26

Area: T = 450009.9999917
Perimeter: p = 1024.26
Semiperimeter: s = 512.13

Angle ∠ A = α = 45.00105494665° = 45°2″ = 0.78554077534 rad
Angle ∠ B = β = 45.00105494665° = 45°2″ = 0.78554077534 rad
Angle ∠ C = γ = 89.99989010669° = 89°59'56″ = 1.57107771468 rad

Height: ha = 3009.999999945
Height: hb = 3009.999999945
Height: hc = 212.1344068692

Median: ma = 335.4087623348
Median: mb = 335.4087623348
Median: mc = 212.1344068692

Inradius: r = 87.86883146696
Circumradius: R = 212.1330000039

Vertex coordinates: A[424.26; 0] B[0; 0] C[212.13; 212.1344068692]
Centroid: CG[212.13; 70.71113562308]
Coordinates of the circumscribed circle: U[212.13; 0.00440686534]
Coordinates of the inscribed circle: I[212.13; 87.86883146696]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 134.9999450533° = 134°59'58″ = 0.78554077534 rad
∠ B' = β' = 134.9999450533° = 134°59'58″ = 0.78554077534 rad
∠ C' = γ' = 90.00110989331° = 90°4″ = 1.57107771468 rad

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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 = 300 ; ; b = 300 ; ; c = 424.26 ; ;

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

p = a+b+c = 300+300+424.26 = 1024.26 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 1024.26 }{ 2 } = 512.13 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 512.13 * (512.13-300)(512.13-300)(512.13-424.26) } ; ; T = sqrt{ 2024999999.26 } = 45000 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 45000 }{ 300 } = 300 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 45000 }{ 300 } = 300 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 45000 }{ 424.26 } = 212.13 ; ;

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{ b**2+c**2-a**2 }{ 2bc } ) = arccos( fraction{ 300**2+424.26**2-300**2 }{ 2 * 300 * 424.26 } ) = 45° 2" ; ; b**2 = a**2+c**2 - 2ac cos beta ; ; beta = arccos( fraction{ a**2+c**2-b**2 }{ 2ac } ) = arccos( fraction{ 300**2+424.26**2-300**2 }{ 2 * 300 * 424.26 } ) = 45° 2" ; ;
 gamma = 180° - alpha - beta = 180° - 45° 2" - 45° 2" = 89° 59'56" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 45000 }{ 512.13 } = 87.87 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin alpha } = fraction{ 300 }{ 2 * sin 45° 2" } = 212.13 ; ;

8. Calculation of medians

m_a = fraction{ sqrt{ 2 b**2+2c**2 - a**2 } }{ 2 } = fraction{ sqrt{ 2 * 300**2+2 * 424.26**2 - 300**2 } }{ 2 } = 335.408 ; ; m_b = fraction{ sqrt{ 2 c**2+2a**2 - b**2 } }{ 2 } = fraction{ sqrt{ 2 * 424.26**2+2 * 300**2 - 300**2 } }{ 2 } = 335.408 ; ; m_c = fraction{ sqrt{ 2 b**2+2a**2 - c**2 } }{ 2 } = fraction{ sqrt{ 2 * 300**2+2 * 300**2 - 424.26**2 } }{ 2 } = 212.134 ; ;
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