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The longest side of a triangle is 3 times the shortest side and the third side is $\mathbf{2} \mathbf{~ c m}$ shorter than the longest side. If the perimeter of the triangle is at least $61 \mathrm{~cm}$, find the minimum length of the shortest side.
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Let shortest side measure $x \mathrm{~cm}$. Therefore the longest side will be $3 x \mathrm{~cm}$ and third side will be $(3 x-2) \mathrm{cm}$ According to the problem,
$x+3 x+3 x-2 \geq 61$
$\Rightarrow \quad 7 x-2 \geq 61$ or $7 x \geq 63$
$\Rightarrow \quad x \geq 9 \mathrm{~cm}$
Hence, the minimum length of the shortest side is $9 \mathrm{~cm}$ and the other sides measure $27 \mathrm{~cm}$ and $25 \mathrm{~cm}$.
$x+3 x+3 x-2 \geq 61$
$\Rightarrow \quad 7 x-2 \geq 61$ or $7 x \geq 63$
$\Rightarrow \quad x \geq 9 \mathrm{~cm}$
Hence, the minimum length of the shortest side is $9 \mathrm{~cm}$ and the other sides measure $27 \mathrm{~cm}$ and $25 \mathrm{~cm}$.
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