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ABCD is a square of side L. A line drawn parallel to the diagonal BD, at a perpendicular distance x from vertex A, cuts the two sides meeting at A. Expressing the area of the part of the square cut off on the A-side of this line as a function of x, find this area at x = 1/sqrt(2) and at x = 2, taking L = 2.

1. 1/2 at x = 1/sqrt(2) and 8*(sqrt(2) - 1) at x = 2
2. 1/2 at x = 1/sqrt(2) and 4*(sqrt(2) - 1) at x = 2
3. 1 at x = 1/sqrt(2) and 8*(sqrt(2) - 1) at x = 2
4. 1/2 at x = 1/sqrt(2) and 2*(sqrt(2) - 1) at x = 2

Correct answer: 1/2 at x = 1/sqrt(2) and 8*(sqrt(2) - 1) at x = 2

Solution

A line parallel to BD at distance x from A cuts off an isosceles right triangle whose distance from the right-angle vertex to the hypotenuse is x, giving legs x*sqrt(2) and area x². Past the centre line, the area is the full square minus the analogous triangle near C.

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