12 17 18 triangle

Acute scalene triangle.

Sides: a = 12   b = 17   c = 18

Area: T = 98.29326116247
Perimeter: p = 47
Semiperimeter: s = 23.5

Angle ∠ A = α = 39.9743659556° = 39°58'25″ = 0.69876719733 rad
Angle ∠ B = β = 65.52114938607° = 65°31'17″ = 1.14435657987 rad
Angle ∠ C = γ = 74.50548465832° = 74°30'17″ = 1.33003548816 rad

Height: ha = 16.38221019374
Height: hb = 11.56438366617
Height: hc = 10.92114012916

Median: ma = 16.44768842034
Median: mb = 12.7188097342
Median: mc = 11.64404467268

Inradius: r = 4.18326643245
Circumradius: R = 9.33994608692

Vertex coordinates: A[18; 0] B[0; 0] C[4.97222222222; 10.92114012916]
Centroid: CG[7.65774074074; 3.64404670972]
Coordinates of the circumscribed circle: U[9; 2.49551010655]
Coordinates of the inscribed circle: I[6.5; 4.18326643245]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 140.0266340444° = 140°1'35″ = 0.69876719733 rad
∠ B' = β' = 114.4798506139° = 114°28'43″ = 1.14435657987 rad
∠ C' = γ' = 105.4955153417° = 105°29'43″ = 1.33003548816 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 = 12 ; ; b = 17 ; ; c = 18 ; ;

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

p = a+b+c = 12+17+18 = 47 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 47 }{ 2 } = 23.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 23.5 * (23.5-12)(23.5-17)(23.5-18) } ; ; T = sqrt{ 9661.44 } = 98.29 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 98.29 }{ 12 } = 16.38 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 98.29 }{ 17 } = 11.56 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 98.29 }{ 18 } = 10.92 ; ;

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{ 12**2-17**2-18**2 }{ 2 * 17 * 18 } ) = 39° 58'25" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 17**2-12**2-18**2 }{ 2 * 12 * 18 } ) = 65° 31'17" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 18**2-12**2-17**2 }{ 2 * 17 * 12 } ) = 74° 30'17" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 98.29 }{ 23.5 } = 4.18 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 12 }{ 2 * sin 39° 58'25" } = 9.34 ; ;




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

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