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(B) Let $a, b, c$ be the sides of the given triangle and $s$ be its semi-perimeter.Then,$s = \frac{a+b+c}{2}$,which implies $2s = a+b+c$ $...(1)$.The

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$4:1$

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(B) Let $a, b, c$ be the sides of the given triangle and $s$ be its semi-perimeter.Then,$s = \frac{a+b+c}{2}$,which implies $2s = a+b+c$ $...(1)$.The area of the given triangle is $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.According to the problem,the sides of the new triangle are $2a, 2b$,and $2c$.Let $S$ be the semi-perimeter of the new triangle.$S = \frac{2a+2b+2c}{2} = a+b+c = 2s$ $...(2)$.The area of the new triangle is $\Delta' = \sqrt{S(S-2a)(S-2b)(S-2c)}$.Substituting $S = 2s$ into the formula:$\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{16s(s-a)(s-b)(s-c)}$$\Delta' = 4\sqrt{s(s-a)(s-b)(s-c)} = 4\Delta$.Thus,the ratio of the area of the new triangle to the given triangle is $\frac{\Delta'}{\Delta} = \frac{4\Delta}{\Delta} = 4:1$.

If each side of a triangle is doubled,then find the ratio of the area of the new triangle thus formed to the area of the given triangle.

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