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(D) In the equilateral triangle $PQR,$ the height $OR$ is given by the Pythagorean theorem in $	riangle OPR$ as:$(OR)^2 = (PR)^2 - (OP)^2$Since $OP =

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$\alpha_1 = 4\alpha_2$

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(D) In the equilateral triangle $PQR,$ the height $OR$ is given by the Pythagorean theorem in $	riangle OPR$ as:$(OR)^2 = (PR)^2 - (OP)^2$Since $OP = l/2,$ we have $(OR)^2 = l^2 - (l/2)^2 = 3l^2/4.$When the temperature changes by $\Delta t,$ the lengths change as $l' = l(1 + \alpha \Delta t).$Let $l_1 = l(1 + \alpha_1 \Delta t)$ be the new length of $PQ,$ so $OP' = \frac{l}{2}(1 + \alpha_1 \Delta t).$Let $l_2 = l(1 + \alpha_2 \Delta t)$ be the new length of $PR,$ so $PR' = l(1 + \alpha_2 \Delta t).$Since $OR$ remains constant,$(OR)^2 = (PR')^2 - (OP')^2.$Equating the initial and final values of $(OR)^2$:$l^2 - (l/2)^2 = [l(1 + \alpha_2 \Delta t)]^2 - [\frac{l}{2}(1 + \alpha_1 \Delta t)]^2$$l^2 - \frac{l^2}{4} = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$Neglecting higher-order terms like $\alpha^2 \Delta t^2$:$l^2 - \frac{l^2}{4} = l^2 + 2l^2 \alpha_2 \Delta t - \frac{l^2}{4} - \frac{l^2}{2} \alpha_1 \Delta t$$0 = 2l^2 \alpha_2 \Delta t - \frac{l^2}{2} \alpha_1 \Delta t$$2 \alpha_2 = \frac{1}{2} \alpha_1 \implies \alpha_1 = 4 \alpha_2.$

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