18 18 25 triangle

Acute isosceles triangle.

Sides: a = 18   b = 18   c = 25

Area: T = 161.8987923087
Perimeter: p = 61
Semiperimeter: s = 30.5

Angle ∠ A = α = 46.01770368696° = 46°1'1″ = 0.80331488054 rad
Angle ∠ B = β = 46.01770368696° = 46°1'1″ = 0.80331488054 rad
Angle ∠ C = γ = 87.96659262607° = 87°57'57″ = 1.53552950428 rad

Height: ha = 17.98986581208
Height: hb = 17.98986581208
Height: hc = 12.9521833847

Median: ma = 19.83768344249
Median: mb = 19.83768344249
Median: mc = 12.9521833847

Inradius: r = 5.30881286258
Circumradius: R = 12.5087881271

Vertex coordinates: A[25; 0] B[0; 0] C[12.5; 12.9521833847]
Centroid: CG[12.5; 4.3177277949]
Coordinates of the circumscribed circle: U[12.5; 0.4443952576]
Coordinates of the inscribed circle: I[12.5; 5.30881286258]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 133.983296313° = 133°58'59″ = 0.80331488054 rad
∠ B' = β' = 133.983296313° = 133°58'59″ = 0.80331488054 rad
∠ C' = γ' = 92.03440737393° = 92°2'3″ = 1.53552950428 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 = 18 ; ; b = 18 ; ; c = 25 ; ;

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

p = a+b+c = 18+18+25 = 61 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 61 }{ 2 } = 30.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 30.5 * (30.5-18)(30.5-18)(30.5-25) } ; ; T = sqrt{ 26210.94 } = 161.9 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 161.9 }{ 18 } = 17.99 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 161.9 }{ 18 } = 17.99 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 161.9 }{ 25 } = 12.95 ; ;

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{ 18**2-18**2-25**2 }{ 2 * 18 * 25 } ) = 46° 1'1" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 18**2-18**2-25**2 }{ 2 * 18 * 25 } ) = 46° 1'1" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-18**2-18**2 }{ 2 * 18 * 18 } ) = 87° 57'57" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 161.9 }{ 30.5 } = 5.31 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 18 }{ 2 * sin 46° 1'1" } = 12.51 ; ;




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

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