22 28 28 triangle

Acute isosceles triangle.

Sides: a = 22 b = 28 c = 28

Area: T = 283.2376650171
Perimeter: p = 78
Semiperimeter: s = 39

Angle ∠ A = α = 46.26547927986° = 46°15'53″ = 0.80774729621 rad
Angle ∠ B = β = 66.86876036007° = 66°52'3″ = 1.16770598458 rad
Angle ∠ C = γ = 66.86876036007° = 66°52'3″ = 1.16770598458 rad

Height: h_a = 25.74987863792
Height: h_b = 20.23111892979
Height: h_c = 20.23111892979

Median: m_a = 25.74987863792
Median: m_b = 20.92884495365
Median: m_c = 20.92884495365

Inradius: r = 7.26224782095
Circumradius: R = 15.22440184927

Vertex coordinates: A[28; 0] B[0; 0] C[8.64328571429; 20.23111892979]
Centroid: CG[12.21442857143; 6.7443729766]
Coordinates of the circumscribed circle: U[14; 5.98108644078]
Coordinates of the inscribed circle: I[11; 7.26224782095]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 133.7355207201° = 133°44'7″ = 0.80774729621 rad
∠ B' = β' = 113.1322396399° = 113°7'57″ = 1.16770598458 rad
∠ C' = γ' = 113.1322396399° = 113°7'57″ = 1.16770598458 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.

$a = 22 ; ; b = 28 ; ; c = 28 ; ;$

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

$p = a+b+c = 22+28+28 = 78 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 78 }{ 2 } = 39 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 39 * (39-22)(39-28)(39-28) } ; ; T = sqrt{ 80223 } = 283.24 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 283.24 }{ 22 } = 25.75 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 283.24 }{ 28 } = 20.23 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 283.24 }{ 28 } = 20.23 ; ;$

5. Calculation of the inner angles of the triangle using a Law of Cosines

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 22**2-28**2-28**2 }{ 2 * 28 * 28 } ) = 46° 15'53" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 28**2-22**2-28**2 }{ 2 * 22 * 28 } ) = 66° 52'3" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-22**2-28**2 }{ 2 * 28 * 22 } ) = 66° 52'3" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 283.24 }{ 39 } = 7.26 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 22 }{ 2 * sin 46° 15'53" } = 15.22 ; ;$

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

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