Two squares are positioned, as shown. The smaller square has side length 1 and the larger square has side length 7. The length of $A B$ is
(A) 14
(D) $\sqrt{85}$
(B) $\sqrt{113}$
(C) 10
(E) $\sqrt{72}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-092.jpg?height=343&width=375&top_left_y=2284&top_left_x=1360)
Two squares are positioned, as shown. The smaller square has side length 1 and the larger square has side length 7 . The length of $A B$ is
(A) 14
(B) $\sqrt{113}$
(C) 10
(D) $\sqrt{85}$
(E) $\sqrt{72}$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-413.jpg?height=352&width=390&top_left_y=1361&top_left_x=1388)

## Solution

From $A$, along the side of the small square, we extend side $A D$ so that it meets the big square at $C$ as shown.

The length of $A C$ is the sum of the side lengths of the squares, i.e. $7+1=8$.

The length of $B C$ is the difference of the side lengths so $B C=6$.

By Pythagoras, $A B^{2}=8^{2}+6^{2}=100, A B=10$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-413.jpg?height=393&width=420&top_left_y=1793&top_left_x=1378)

ANSWER: (C)