Area formula: S=√[p(p-a)(p-b)(p-c)]? And p in the formula is a half circumference, that is, p=(a+b+c)/2, so s = √ [12 (12-7) (12-8)] = √.

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