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(B) Let the initial length of each rod be $l$. In the equilateral triangle $ABC$,$D$ is the mid-point of $AB$,so $AD = l/2$. The length of the altitud

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$\alpha_1 = 4\alpha_2$

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(B) Let the initial length of each rod be $l$. In the equilateral triangle $ABC$,$D$ is the mid-point of $AB$,so $AD = l/2$. The length of the altitude $DC$ is given by $DC^2 = AC^2 - AD^2 = l^2 - (l/2)^2 = 3l^2/4$.After a small change in temperature $\Delta t$,the new lengths become $l' = l(1 + \alpha \Delta t)$.The new length of $AC$ is $l_{AC}' = l(1 + \alpha_2 \Delta t)$ and the new length of $AD$ is $l_{AD}' = (l/2)(1 + \alpha_1 \Delta t)$.Since $DC$ remains constant,$DC^2 = (l_{AC}')^2 - (l_{AD}')^2$.$3l^2/4 = [l(1 + \alpha_2 \Delta t)]^2 - [(l/2)(1 + \alpha_1 \Delta t)]^2$.$3l^2/4 = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - (l^2/4)(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.Neglecting higher-order terms $\alpha^2 \Delta t^2$,we get:$3l^2/4 = l^2 + 2l^2 \alpha_2 \Delta t - l^2/4 - (l^2/4)(2\alpha_1 \Delta t)$.$3l^2/4 = 3l^2/4 + 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.$0 = 2\alpha_2 \Delta t - (1/2)\alpha_1 \Delta t$.$2\alpha_2 = \alpha_1/2 \implies \alpha_1 = 4\alpha_2$.

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