Math Forum: Ask Dr. Math FAQ: Triangle Formulas

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also see Defining Geometric Figures

Triangle

A polygon (plane figure) with
3 angles and 3 sides.
Sides: a, b, c
Opposite angles: A, B, C

Altitudes: h_a, h_b, h_c
Medians: m_a , m_b , m_c
Angle bisectors: t_a, t_b, t_c

Perimeter: P
Semiperimeter: s
Area: K

Radius of circumscribed circle: R
Radius of inscribed circle: r

To read about triangles, visit The Geometry Center.

Equilateral Triangle

A triangle with all three sides of equal length.

a = b = c.
A = B = C = Pi/3 radians = 60^o

P = 3a
s = 3a/2
K = a²sqrt(3)/4

h_a = m_a = t_a = a sqrt(3)/2

R = a sqrt(3)/3
r = a sqrt(3)/6

JavaSketchpad exploration: Equilateral triangle

Isosceles Triangle
	A triangle with two sides of equal length. a = c A = C
		B + 2A = Pi radians = 180^o P = 2a + b s = a + b/2 K = b sqrt(4a²-b²)/4 = a² sin(B)/2 = ab sin(A)/2 h_a = b sqrt(4a²-b²)/(2a) = a sin(B) = b sin(A) m_a = sqrt(a²+2b²)/2 t_a = b sqrt(a[2a+b])/(a+b) = b sin(A)/sin(3A/2) h_b = m_b = t_b = sqrt(4a²-b²)/2 = a cos(B/2) R = a²b/4K = a/[2 sin(A)] = b/[2 sin(B)] r = K/s = b sqrt[(2a-b)/(2a+b)]/2
		JavaSketchpad exploration: Isosceles triangle

Right Triangle
	A triangle with one right angle. C = A + B = Pi/2 radians = 90^oc² = a² + b² (Pythagorean Theorem)
		P = a + b + c s = (a+b+c)/2 K = ab/2 h_a = b h_b = a h_c = ab/c m_a = sqrt(4b²+a²)/2 m_b = sqrt(4a²+b²)/2 m_c = c/2
		t_a = 2bc cos(A/2)/(b+c) = sqrt[bc(1-a²/[b+c]²)] t_b = 2ac cos(B/2)/(a+c) = sqrt[ac(1-b²/[a+c]²)] t_c = ab sqrt(2)/(a+b) R = c/2 r = ab/(a+b+c) = s - c
		JavaSketchpad exploration: Right triangle

Scalene Triangle

	A triangle with no two sides equal. (Note that the following formulas work with all triangles, not just scalene triangles.) P = a + b + c s = (a+b+c)/2 K = ah_a/2 = ab sin(C)/2 = a² sin(B) sin(C)/[2 sin(A)] = sqrt[s(s-a)(s-b)(s-c)] (Heron's or Hero's Formula) h_a = c sin(B) = b sin(C) = 2K/a m_a = sqrt(2b²+2c²-a²)/2 t_a = 2bc cos(A/2)/(b+c) = sqrt[bc(1-a²/[b+c]²)]
	R = abc/4K = a/[2 sin(A)] = b/[2 sin(B)] = c/[2 sin(C)] r = 2K/P = K/s = sqrt[(s-a)(s-b)(s-c)/s] = c sin(A/2)sin(B/2)/cos(C/2) = ab sin(C)/(2 s) = (s-c)tan(C/2)
	JavaSketchpad exploration: General triangle

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