Discussion

Explanation:

The sides of the rectangle PRSU are PR = 15 cm and RS = 10 cm.
∴ Its diagonal = $\sqrt{{15}^2+{10}^2}=\sqrt{325}\approx 18$ cm.

Given that the minimum distance between any pair of points formed by taking one from A, B and F and the other from C, D and E is $10\sqrt{3}=\sqrt{300}\approx 17$ cm, which is close to the length of the diagonal of PRSU. This implies that each of A, B and F are close to one of the vertices P or U and each of C, D and E are close to one of the vertices S or R.
Note that all the three points in one of the rectangles PQTU and QRST will be close to the same vertex (or else the minimum distance between one of these points and one of the three in the other rectangle will be less than $10\sqrt{3}$ cm). Also, the points (A, B, F) and (C, D, E) have to be diagonally opposite as maximum distance between two vertices on same side is 15 cm.

Option (a): The closest two points of the given six points have to be any two out of A, B, F or any two out of C, D, E (since they are closest to the same vertex). Therefore, (F, C) cannot be the closest pair of points as they are diagonally opposite.

Option (b): It is definitely false as A and B are close to the same vertex, while F and C are close to diagonally opposite vertices.

Option (c): It is possible but not necessary that the closest pair of points among the six given points is (C,D), (D, E) or (C, E). The other possibilities are (A, B), (B,F) or (A,F).

Hence, option (a).

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