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A rectangular box is rested on a horizontal surface. Length, breadth and height of the box are $$20\text{ cm}$$, $$12\text{ cm}$$ and $$9\text{ cm}$$ respectively. The box has 5 faces; it is open from the top. Its vertices are named $$A, B, C, D, P, Q, R, S$$ as shown in the figure. There are two ants; ant-1 at the vertex $$A$$ and ant-2 at the vertex $$Q$$ inside the box at rest. Neither ant can fly, but only can crawl along the face or edge of the box. Ant-1 has to reach at the vertex $$R$$ going along the edges or a face of the box from outside of the box and ant-2 has to reach vertex $$D$$ going from inside the box. Both ants start at the same time at the same speed of $$1\text{ cm/s}$$, and both follow the shortest possible route. Find the time taken by both the ants to reach their respective destinations.
Correct Answer: 62.24
Here $$AD = PS = 12$$, $$AB = PQ = 20$$ and the height is $$AP = 9$$. Ant-1 crawls outside, so unfold the left face $$APSD$$ and the back face $$DSRC$$ into one $$32 \times 9$$ rectangle (the front and right faces give the same); its shortest route is the diagonal $$AR = \sqrt{32^2 + 9^2} = \sqrt{1105} \approx 33.24\text{ cm}$$, so $$t_1 = \sqrt{1105} \approx 33.24$$ seconds. Ant-2 crawls inside, so unfold the back wall $$SRCD$$ onto the base $$PQRS$$ to get a $$20 \times 21$$ rectangle, giving $$DQ = \sqrt{20^2 + 21^2} = \sqrt{841} = 29\text{ cm}$$ (the other unfoldings give the longer $$\sqrt{985}$$ and $$\sqrt{1105}$$), so $$t_2 = 29$$ seconds. Together $$t_1 + t_2 = \sqrt{1105} + 29 \approx 62.24$$ seconds.
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