As shown in the figure,an equilateral triangl | vedclass

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(B) Let the length of each rod be $l$. In the right-angled triangle $ADC$,by Pythagoras theorem:$DC^2 = AC^2 - AD^2$Since $D$ is the midpoint of $AB$,

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$\alpha_1 = 4\alpha_2$

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(B) Let the length of each rod be $l$. In the right-angled triangle $ADC$,by Pythagoras theorem:$DC^2 = AC^2 - AD^2$Since $D$ is the midpoint of $AB$,$AD = \frac{l}{2}$.Thus,$DC^2 = l^2 - (\frac{l}{2})^2 = l^2 - \frac{l^2}{4} = \frac{3l^2}{4}$.When the temperature increases by $\Delta T$,the new lengths are $l' = l(1 + \alpha \Delta T)$.$AC' = l(1 + \alpha_2 \Delta T)$ and $AD' = \frac{l}{2}(1 + \alpha_1 \Delta T)$.The new length $DC'$ is given by:$DC'^2 = AC'^2 - AD'^2 = [l(1 + \alpha_2 \Delta T)]^2 - [\frac{l}{2}(1 + \alpha_1 \Delta T)]^2$$DC'^2 = l^2(1 + 2\alpha_2 \Delta T + \alpha_2^2 \Delta T^2) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta T + \alpha_1^2 \Delta T^2)$Neglecting higher-order terms of $\Delta T$ (i.e.,$\alpha^2 \Delta T^2 \approx 0$):$DC'^2 \approx l^2(1 + 2\alpha_2 \Delta T) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta T)$$DC'^2 \approx (l^2 - \frac{l^2}{4}) + (2l^2\alpha_2 \Delta T - \frac{l^2}{2}\alpha_1 \Delta T)$For $DC$ to remain constant,the change in $DC^2$ must be zero:$2l^2\alpha_2 \Delta T - \frac{l^2}{2}\alpha_1 \Delta T = 0$$2\alpha_2 = \frac{\alpha_1}{2} \Rightarrow \alpha_1 = 4\alpha_2$.

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