2001 CEMC Gauss (Grade 8) Problems/Problem 18

Problem

Two squares are positioned, as shown. The smaller square has side length $1$ and the larger square has side length $7$. The length of $AB$ is

$\text{ (A) }\ 14 \qquad\text{ (B) }\ \sqrt{113} \qquad\text{ (C) }\ 10 \qquad\text{ (D) }\ \sqrt{85} \qquad\text{ (E) }\ \sqrt{72}$

Solution 1

We can draw point $C$ so that $AC$ and $BC$ are perpendicular at $C$, and then draw $AC$. We can also make $AD$ and $DE$ of the sides of the square, like so:


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


$AD = 1$, $DE = 1$, and $CD = 7$ because these are the side lengths of the square, so $AC = 1 + 7 = 8$ and $BC = 7 - 1 = 6$.


An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.


We then have a right triangle. We have a $6-8-10$ right triangle, so we have $AB = \boxed {\textbf {(C) } 10}$.

~anabel.disher

2001 CEMC Gauss (Grade 8) (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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CEMC Gauss (Grade 8)