2012/13 AHSMC Part 1 Problem 11

Problem 11 of 16

$ABCD$ is a rectangle such that $AD - AB = 15$ cm. $PQRS$ is a square inside $ABCD$ whose sides are parallel to those of the rectangle, with $P$ closest to $A$ and $Q$ closest to $B$ . The total area of $APSD$ and $BQRC$ is $363$ cm $^2$ while the total area of $APQB$ and $CRSD$ is $1113$ cm $^2$ . What, in cm $^2$ , is the area of $PQRS$ ?

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Solution

We shade the regions $APQB$ and $CRSD$ while leaving the regions $APSD$ and $BQRC$ unshaded. Extend the sides of $PQRS$ to the perimeter of $ABCD$ , creating four rectangles at the corners each of which consists of two congruent triangles, one shaded and one unshaded. The difference between the total area of the unshaded regions (not counting $PQRS$ ) and the total area of the shaded regions is $1113 - 363 = 750$ cm $^2$ . The difference in the lengths of $AD$ and $AB$ is $15$ cm. Hence the side length of $PQRS$ is $750 \div 15 = 50$ cm, and the area of $PQRS$ is $2500$ cm $^2$ . The answer is (C).

University of Alberta

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