Question

$A=\sqrt{S(S-a)(S-b)(S-c)}$

Make S the subject of the formula

Answer 1

$A=\sqrt {S(S−a)(S−b)(S−c)}$

square both sides of the equation to get:

$A^2=(\sqrt {S(S−a)(S−b)(S−c)})^2$

$A^2={S(S−a)(S−b)(S−c)}$

but $S={a+b+c\over 2}$

$\implies A^2={{a+b+c\over 2}({a+b+c\over 2}−a)({a+b+c\over 2}−b)({a+b+c\over 2}−c)}$

$={{a+b+c\over 2}({a+b+c-2a\over 2})({a+b+c-2b\over 2})({a+b+c-2c\over 2})}$

$={{a+b+c\over 2}({b+c-a\over 2})({a+c-b\over 2})({a+b-c\over 2})}$

Multiply both sides of the equation by ${2\over a+b+c}$

$\implies A^2 \times {2\over a+b+c}={2\over a+b+c}\times {{a+b+c\over 2}({b+c-a\over 2})({a+c-b\over 2})({a+b-c\over 2})}$

$\implies {2A^2\over a+b+c}={({b+c-a\over 2})({a+c-b\over 2})({a+b-c\over 2})}$

$\implies {2A^2\over a+b+c}={{(b+c-a)(a+c-b)(a+b-c)\over 8}}$

Cross multiply to get

$\implies 16A^2=(a+b+c)(b+c-a)(a+c-b)(a+b-c)$

make $(a+b+c)$ subject of the formula to get

$(a+b+c)={16A^2\over (b+c-a)(a+c-b)(a+b-c)}$

Divide both sides by 2 to get

${(a+b+c)\over 2}={16A^2\over 2 (b+c-a)(a+c-b)(a+b-c)}$

$\implies {(a+b+c)\over 2}={8A^2\over (b+c-a)(a+c-b)(a+b-c)}$

Recall that $S={a+b+c\over 2}$

$\therefore$ $S={8A^2\over (b+c-a)(a+c-b)(a+b-c)}$