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(D) Let the sides of the original triangle be $a$,$b$,and $c$. The semi-perimeter is $s = \frac{a+b+c}{2}$.According to Heron's formula,the area of th

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$ 4 $

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(D) Let the sides of the original triangle be $a$,$b$,and $c$. The semi-perimeter is $s = \frac{a+b+c}{2}$.According to Heron's formula,the area of the original triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.If every side is doubled,the new sides are $2a$,$2b$,and $2c$.The new semi-perimeter $s'$ is $s' = \frac{2a+2b+2c}{2} = 2s$.The new area $\Delta'$ is $\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}$.Substituting the values: $\Delta' = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}$.$\Delta' = \sqrt{2s \cdot 2(s-a) \cdot 2(s-b) \cdot 2(s-c)}$.$\Delta' = \sqrt{16 \cdot s(s-a)(s-b)(s-c)}$.$\Delta' = 4 \sqrt{s(s-a)(s-b)(s-c)} = 4\Delta$.Thus,the new area is $4$ times the old area,so $K = 4$.

If every side of a triangle is doubled,the area of the new triangle is $K$ times the area of the old one. $K$ is equal to

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