Geometría matemática de la escuela secundaria

Fórmula del área: S=√[p(p-a)(p-b)(p-c)]? Y p en la fórmula es el medio perímetro, es decir, p=(a b c)/2, entonces s =√[12(12-7)(12-8)(12-9)]=√.

S=(absin∠C)/2

(sin∠C)^2 (cos∠C)^2=1

c^2= a^2 b^2-2abcos∠C

cos∠C=(a^2 b^2-c^2)/(2ab)

sin∠c=√[ 1-(cos∠c)^2]=√{1-[(a^2 b^2-c^2)/(2ab)]^2}

S=(absin∠C) /2

=ab√{1-[(a^2 b^2-c^2)/(2ab)]^2}/2

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

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

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

=√{(ab)^2-[(a^2 b^2-c^2)/2]^2}/ 2

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

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

=√{{[(a b)^2-c^2]/2}{-[(a-b)^2-c^2]/2}}/2

= √ {[(a b c)(a b-c)/2][-(a b c)(a-b-c)/2]}/2

=√[-(a b c)(a b-c)(a-b c)( a-b-c)/4]/2

=√[(a b c)(a b-c)(a-b c)(-a b c)]/4

=√[(a b-c)(a b-c)(a c-b)(b c-a)]/4

=√[(7 8 9)(7 8-9)(7 9-8)(8 9-7)]/4

=√(24×6×8×10)/4

=√(12×2×6×4×2×2×5)/4

=12×2×2×√5/4

=12√5