21 25 25 triangle

Acute isosceles triangle.

Sides: a = 21 b = 25 c = 25

Area: T = 238.2255077395
Perimeter: p = 71
Semiperimeter: s = 35.5

Angle ∠ A = α = 49.66991749794° = 49°40'9″ = 0.86768906401 rad
Angle ∠ B = β = 65.16554125103° = 65°9'55″ = 1.13773510067 rad
Angle ∠ C = γ = 65.16554125103° = 65°9'55″ = 1.13773510067 rad

Height: h_a = 22.68881026091
Height: h_b = 19.05880061916
Height: h_c = 19.05880061916

Median: m_a = 22.68881026091
Median: m_b = 19.41100489438
Median: m_c = 19.41100489438

Inradius: r = 6.71105655604
Circumradius: R = 13.7743738835

Vertex coordinates: A[25; 0] B[0; 0] C[8.82; 19.05880061916]
Centroid: CG[11.27333333333; 6.35326687305]
Coordinates of the circumscribed circle: U[12.5; 5.78549703107]
Coordinates of the inscribed circle: I[10.5; 6.71105655604]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 130.3310825021° = 130°19'51″ = 0.86768906401 rad
∠ B' = β' = 114.835458749° = 114°50'5″ = 1.13773510067 rad
∠ C' = γ' = 114.835458749° = 114°50'5″ = 1.13773510067 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.

$a = 21 ; ; b = 25 ; ; c = 25 ; ;$

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

$p = a+b+c = 21+25+25 = 71 ; ;$

2. Semiperimeter of the triangle

$s = fraction{ o }{ 2 } = fraction{ 71 }{ 2 } = 35.5 ; ;$

3. The triangle area using Heron's formula

$T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 35.5 * (35.5-21)(35.5-25)(35.5-25) } ; ; T = sqrt{ 56751.19 } = 238.23 ; ;$

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

$T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 238.23 }{ 21 } = 22.69 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 238.23 }{ 25 } = 19.06 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 238.23 }{ 25 } = 19.06 ; ;$

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

$a**2 = b**2+c**2 - 2bc cos( alpha ) ; ; alpha = arccos( fraction{ a**2-b**2-c**2 }{ 2bc } ) = arccos( fraction{ 21**2-25**2-25**2 }{ 2 * 25 * 25 } ) = 49° 40'9" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 25**2-21**2-25**2 }{ 2 * 21 * 25 } ) = 65° 9'55" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-21**2-25**2 }{ 2 * 25 * 21 } ) = 65° 9'55" ; ;$

6. Inradius

$T = rs ; ; r = fraction{ T }{ s } = fraction{ 238.23 }{ 35.5 } = 6.71 ; ;$

7. Circumradius

$R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 21 }{ 2 * sin 49° 40'9" } = 13.77 ; ;$

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