2003 CEMC Pascal Problems/Problem 15

Problem

In the diagram, square $ABCD$ is made up of $36$ squares, each with side length $1$ . The area of the square $KLMN$ , in square units, is

$\text{ (A) }\ 12 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 18 \qquad\text{ (D) }\ 20 \qquad\text{ (E) }\ 25$

Solution

Since the side lengths of a square are all equal to each other, we only need to find one of the side lengths of square $KLMN$ .

The angles of squares are also all $90^{\circ}$ , so we know that triangle $KBL$ is a right triangle, which means that we can use the pythagorean theorem.

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$KB = 4$ , since the side lengths of the tiny squares are all $1$ . Using this logic, we also get $BL = 2$ .

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Using the pythagorean theorem, we have:

$KB^2 + BL^2 = KL^2$

$4^2 + 2^2 = KL^2$

$KL^2 = 16 + 4 = 20$

$KL$ is one of the side lengths of square $KLMN$ , and the area of a square is the square of its side length. We can see that $KL^2 = \boxed {\textbf {(D) } 20}$ .

~anabel.disher

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2003 CEMC Pascal Problems/Problem 15

Problem

Solution