A $5 \mathrm{~cm}$ by $5 \mathrm{~cm}$ pegboard and a $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ pegboard each have holes at the intersection of invisible horizontal and vertical lines that occur in $1 \mathrm{~cm}$ intervals from each edge. Pegs are placed into the holes on the two main diagonals of both pegboards. The $5 \mathrm{~cm}$ by $5 \mathrm{~cm}$ pegboard is shown; it has 16 holes. The 8 shaded holes have pegs, and the 8 unshaded holes do not. How many empty holes does the $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ pegboard have?

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-011.jpg?height=355&width=357&top_left_y=581&top_left_x=1315)
A $10 \mathrm{~cm}$ by $10 \mathrm{~cm}$ board has 9 rows of 9 holes, or $9 \times 9=81$ pegs in total.

Each hole on the 2 main diagonals has a peg in it.

There are 9 holes on each diagonal, with the centre hole on both diagonals, since there is an odd number of holes in each row.

Therefore, the total number of holes on the two diagonals is $9+9-1=17$.

This means that the number of empty holes is $81-17=64$.

ANSWER: 64