Question

S=根号下[l(l-a)(l-b)(l-c)]

Answer 1

△ABC中已知三边长分别是a、b、c.依余弦定理

cosC=(a^2+b^2-c^2)/(2ab)

sinC=√[1-(sinC)^2]

=√{1-[(a^2+b^2-c^2)/(2ab)]^2}

=1/(2ab)*√[(2ab)^2-(a^2+b^2-c^2)]

=1/(2ab)√[(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)]

=1/(2ab)√{[(a+b)^2-c^2]*[c^2-(a-b)^2]}

=1/2ab)*√(a+b+c)(a+b-c)(c+a-b)(c-a+b)]

令p=(a+b+c)/2,则p-a=(b+c-a)/2,p-b=(c+a-b)/2,p-c=(a+b-c)/2

所以sinC=1/(2ab)*√[16p(p-c)(p-b)(p-a)]

因而S(△)=(1/2)absinC

=(ab/2)*1/(2ab)*4√[p(p-a)(p-b)(p-c)]

=√[p(p-a)(p-b)(p-c)].