7 15 15 triangle

Acute isosceles triangle.

Sides: a = 7   b = 15   c = 15

Area: T = 51.05108325104
Perimeter: p = 37
Semiperimeter: s = 18.5

Angle ∠ A = α = 26.98767976431° = 26°59'12″ = 0.47110084734 rad
Angle ∠ B = β = 76.50766011784° = 76°30'24″ = 1.33552920901 rad
Angle ∠ C = γ = 76.50766011784° = 76°30'24″ = 1.33552920901 rad

Height: ha = 14.58659521458
Height: hb = 6.8076777668
Height: hc = 6.8076777668

Median: ma = 14.58659521458
Median: mb = 8.98661003778
Median: mc = 8.98661003778

Inradius: r = 2.765950446
Circumradius: R = 7.71329006646

Vertex coordinates: A[15; 0] B[0; 0] C[1.63333333333; 6.8076777668]
Centroid: CG[5.54444444444; 2.26989258893]
Coordinates of the circumscribed circle: U[7.5; 1.87996768218]
Coordinates of the inscribed circle: I[3.5; 2.765950446]

Exterior(or external, outer) angles of the triangle:
∠ A' = α' = 153.0133202357° = 153°48″ = 0.47110084734 rad
∠ B' = β' = 103.4933398822° = 103°29'36″ = 1.33552920901 rad
∠ C' = γ' = 103.4933398822° = 103°29'36″ = 1.33552920901 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 = 7 ; ; b = 15 ; ; c = 15 ; ;

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

p = a+b+c = 7+15+15 = 37 ; ;

2. Semiperimeter of the triangle

s = fraction{ o }{ 2 } = fraction{ 37 }{ 2 } = 18.5 ; ;

3. The triangle area using Heron's formula

T = sqrt{ s(s-a)(s-b)(s-c) } ; ; T = sqrt{ 18.5 * (18.5-7)(18.5-15)(18.5-15) } ; ; T = sqrt{ 2606.19 } = 51.05 ; ;

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

T = fraction{ a h _a }{ 2 } ; ; h _a = fraction{ 2 T }{ a } = fraction{ 2 * 51.05 }{ 7 } = 14.59 ; ; h _b = fraction{ 2 T }{ b } = fraction{ 2 * 51.05 }{ 15 } = 6.81 ; ; h _c = fraction{ 2 T }{ c } = fraction{ 2 * 51.05 }{ 15 } = 6.81 ; ;

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{ 7**2-15**2-15**2 }{ 2 * 15 * 15 } ) = 26° 59'12" ; ; beta = arccos( fraction{ b**2-a**2-c**2 }{ 2ac } ) = arccos( fraction{ 15**2-7**2-15**2 }{ 2 * 7 * 15 } ) = 76° 30'24" ; ; gamma = arccos( fraction{ c**2-a**2-b**2 }{ 2ba } ) = arccos( fraction{ 15**2-7**2-15**2 }{ 2 * 15 * 7 } ) = 76° 30'24" ; ;

6. Inradius

T = rs ; ; r = fraction{ T }{ s } = fraction{ 51.05 }{ 18.5 } = 2.76 ; ;

7. Circumradius

R = fraction{ a }{ 2 * sin( alpha ) } = fraction{ 7 }{ 2 * sin 26° 59'12" } = 7.71 ; ;




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

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