10 24 28 triangle

Obtuse scalene triangle.

Sides: a = 10   b = 24   c = 28

Area: T = 116.9233051619
Perimeter: p = 62
Semiperimeter: s = 31

Angle ∠ A = α = 20.36441348063° = 20°21'51″ = 0.35554212017 rad
Angle ∠ B = β = 56.63329870308° = 56°37'59″ = 0.98884320889 rad
Angle ∠ C = γ = 103.0032878163° = 103°10″ = 1.7987739363 rad

Height: ha = 23.38546103239
Height: hb = 9.7443587635
Height: hc = 8.35216465442

Median: ma = 25.59329677841
Median: mb = 17.26326765016
Median: mc = 11.91663752878

Inradius: r = 3.77217113426
Circumradius: R = 14.36884241621

Vertex coordinates: A[28; 0] B[0; 0] C[5.5; 8.35216465442]
Centroid: CG[11.16766666667; 2.78438821814]
Coordinates of the circumscribed circle: U[14; -3.23328954365]
Coordinates of the inscribed circle: I[7; 3.77217113426]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 159.6365865194° = 159°38'9″ = 0.35554212017 rad
∠ B' = β' = 123.3677012969° = 123°22'1″ = 0.98884320889 rad
∠ C' = γ' = 76.99771218371° = 76°59'50″ = 1.7987739363 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 = 10 ; ; b = 24 ; ; c = 28 ; ;

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

p = a+b+c = 10+24+28 = 62 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 62 }{ 2 } = 31 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 31 * (31-10)(31-24)(31-28) } ; ; T = sqrt{ 13671 } = 116.92 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 116.92 }{ 10 } = 23.38 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 116.92 }{ 24 } = 9.74 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 116.92 }{ 28 } = 8.35 ; ;

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{ 10**2-24**2-28**2 }{ 2 * 24 * 28 } ) = 20° 21'51" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-10**2-28**2 }{ 2 * 10 * 28 } ) = 56° 37'59" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 28**2-10**2-24**2 }{ 2 * 24 * 10 } ) = 103° 10" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 116.92 }{ 31 } = 3.77 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 10 }{ 2 * sin 20° 21'51" } = 14.37 ; ;




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

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