If the sides of a triangle are doubled,its ar | vedclass

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(C) Let the original sides of the triangle be $a, b,$ and $c$. The semi-perimeter $s$ is given by $s = \frac{1}{2}(a + b + c)$. The area of the origin

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becomes $4$ times

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(C) Let the original sides of the triangle be $a, b,$ and $c$. The semi-perimeter $s$ is given by $s = \frac{1}{2}(a + b + c)$. The area of the original triangle is $A = \sqrt{s(s - a)(s - b)(s - c)}$. If the sides are doubled,the new sides are $2a, 2b,$ and $2c$. The new semi-perimeter $S$ is $S = \frac{1}{2}(2a + 2b + 2c) = a + b + c = 2s$. The area of the new triangle $A'$ is $\sqrt{S(S - 2a)(S - 2b)(S - 2c)}$. Substituting $S = 2s$,we get $A' = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}$. Factoring out $2$ from each term inside the square root,we get $A' = \sqrt{2s \cdot 2(s - a) \cdot 2(s - b) \cdot 2(s - c)} = \sqrt{16s(s - a)(s - b)(s - c)}$. $A' = 4 \sqrt{s(s - a)(s - b)(s - c)} = 4A$. Thus,the area becomes $4$ times the original area.

If the sides of a triangle are doubled,its area

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