Question

ABCD is a square. AM=9. What is the measure of BD?

(1 point)

Answer 1

To solve this problem, we start by establishing some basic information about the square ABCD.

Let the vertices of the square be:
- \( A \) at \( (0, 0) \)
- \( B \) at \( (s, 0) \)
- \( C \) at \( (s, s) \)
- \( D \) at \( (0, s) \)

where \( s \) is the length of a side of the square.

The point \( M \) is on side \( AB \), which means that \( M \) can be represented by the coordinates \( (x, 0) \) where \( 0 \leq x \leq s \). We know \( AM = 9 \), which implies \( x = 9 \).

However, \( x \) must be less than or equal to \( s \) (the side length of the square), so for the square to exist, \( s \) must be at least 9. Therefore, we can assume \( s = 9 \) or \( s > 9 \).

Next, we calculate the length of the diagonal \( BD \).

The coordinates of point \( B \) are \( (s, 0) \) and the coordinates of point \( D \) are \( (0, s) \).

The length of diagonal \( BD \) can be found using the distance formula:
\[
BD = \sqrt{(s - 0)^2 + (0 - s)^2} = \sqrt{s^2 + s^2} = \sqrt{2s^2} = s\sqrt{2}
\]

If we set \( s = 9 \):
\[
BD = 9\sqrt{2}
\]

However, if \( s \) is greater than 9, \( BD \) is simply based on the size of the square and will still be \( s\sqrt{2} \).

Thus, in either case, the measure of diagonal \( BD \) when \( AM = 9 \) is:

\[
\text{Length of } BD = s\sqrt{2}
\]

If \( s = 9 \):

\[
BD = 9\sqrt{2}
\]

Thus, the length of diagonal \( BD \) is \( 9\sqrt{2} \).