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(C) Let the original sides of the triangle be $a, b,$ and $c$. The original area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where

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$800$

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(C) Let the original sides of the triangle be $a, b,$ and $c$. The original area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.When each side is increased by $200 \%$,the new sides become $a' = a + 2a = 3a$,$b' = b + 2b = 3b$,and $c' = c + 2c = 3c$.The new semi-perimeter $s'$ is $s' = \frac{3a+3b+3c}{2} = 3 \left( \frac{a+b+c}{2} ight) = 3s$.The new area $A'$ is $A' = \sqrt{s'(s'-a')(s'-b')(s'-c')} = \sqrt{3s(3s-3a)(3s-3b)(3s-3c)}$.Factoring out $3$ from each term inside the square root: $A' = \sqrt{3s \cdot 3(s-a) \cdot 3(s-b) \cdot 3(s-c)} = \sqrt{81 \cdot s(s-a)(s-b)(s-c)} = 9 \sqrt{s(s-a)(s-b)(s-c)} = 9A$.The increase in area is $A' - A = 9A - A = 8A$.The percentage increase is $\frac{8A}{A} 	imes 100 \% = 800 \%$.

If all the sides of a triangle are increased by $200 \%$,then the area of the triangle will increase by (in $\%$):

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