30:T76b,Then area becomes, A₂ =

{{{A_1}} \over 3}

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\begin{aligned} & R_1=\frac{p l_1}{A_1} \\\\ & R_2=\frac{p l_2}{A_2}=\frac{p \times 3 l_1}{A_1 / 3}=9 \frac{p l_1}{A_1}=9 R_1 \\\\ & \therefore R_2: R_1=9: 1 \end{aligned}

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