15 24 25 triangle

Acute scalene triangle.

Sides: a = 15   b = 24   c = 25

Area: T = 174.5399393834
Perimeter: p = 64
Semiperimeter: s = 32

Angle ∠ A = α = 35.57771025511° = 35°34'38″ = 0.62109375778 rad
Angle ∠ B = β = 68.57219019579° = 68°34'19″ = 1.19768054635 rad
Angle ∠ C = γ = 75.85109954911° = 75°51'4″ = 1.32438496122 rad

Height: ha = 23.27219191779
Height: hb = 14.54549494862
Height: hc = 13.96331515067

Median: ma = 23.32991662946
Median: mb = 16.76330546142
Median: mc = 15.62884996081

Inradius: r = 5.45443560573
Circumradius: R = 12.89110726145

Vertex coordinates: A[25; 0] B[0; 0] C[5.48; 13.96331515067]
Centroid: CG[10.16; 4.65443838356]
Coordinates of the circumscribed circle: U[12.5; 3.15111510835]
Coordinates of the inscribed circle: I[8; 5.45443560573]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 144.4232897449° = 144°25'22″ = 0.62109375778 rad
∠ B' = β' = 111.4288098042° = 111°25'41″ = 1.19768054635 rad
∠ C' = γ' = 104.1499004509° = 104°8'56″ = 1.32438496122 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 = 15 ; ; b = 24 ; ; c = 25 ; ;

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

p = a+b+c = 15+24+25 = 64 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 64 }{ 2 } = 32 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 32 * (32-15)(32-24)(32-25) } ; ; T = sqrt{ 30464 } = 174.54 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 174.54 }{ 15 } = 23.27 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 174.54 }{ 24 } = 14.54 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 174.54 }{ 25 } = 13.96 ; ;

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{ 15**2-24**2-25**2 }{ 2 * 24 * 25 } ) = 35° 34'38" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 24**2-15**2-25**2 }{ 2 * 15 * 25 } ) = 68° 34'19" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 25**2-15**2-24**2 }{ 2 * 24 * 15 } ) = 75° 51'4" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 174.54 }{ 32 } = 5.45 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 15 }{ 2 * sin 35° 34'38" } = 12.89 ; ;




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

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