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A rectangular plank, with a width of sqrt(2) meters, is positioned symmetrically on the diagonal of an 8-meter side square. What is the area of this plank?
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Solution: Step 1: Calculate the length of the diagonal of the square. Diagonal AC = side * sqrt(2) = 8 * sqrt(2) meters.
Step 2: Let 'x' be the side length of the small isosceles right triangles formed at each corner of the square where the plank begins (e.g., AP = AQ = x). The width of the plank, PQ, forms the hypotenuse of such a triangle. So, PQ = sqrt(x^2 + x^2) = x * sqrt(2).
Step 3: Given the width of the plank is sqrt(2) meters, set x * sqrt(2) = sqrt(2). This gives x = 1 meter.
Step 4: The length along the square's diagonal, from a corner (A) to the point where the plank starts (T, as per solution's implicit diagram), is the altitude of the isosceles right triangle APQ, which is AT = x / sqrt(2) = 1 / sqrt(2) meters.
Step 5: The effective length of the plank along the diagonal is the total diagonal length minus these two end segments. Length of plank (L) = AC - 2 * AT = 8 * sqrt(2) - 2 * (1 / sqrt(2)).
Step 6: Simplify the plank's length: L = 8 * sqrt(2) - sqrt(2) = 7 * sqrt(2) meters.
Step 7: Calculate the area of the plank. Area = Length * Width = (7 * sqrt(2)) * (sqrt(2)) = 7 * 2 = 14 sq. meters.