Question

P is the midpoint of $\overline{BD}$. $AP = BP = 4$, $\overline{AP} \perp \overline{BD}$, $\overline{BD} \perp \overline{DC}$, $\overline{AB} \perp \overline{BC}$. In simplest radical form, what is the perimeter of pentagon $ABCDP$?

Answer 1

The person who answered before this is correct!

Answer 2

Let's break down the information given:

1. \(P\) is the midpoint of \(\overline{BD}\).
2. \(AP = BP = 4\).
3. \(\overline{AP} \perp \overline{BD}\).
4. \(\overline{BD} \perp \overline{DC}\).
5. \(\overline{AB} \perp \overline{BC}\).

Since \(P\) is the midpoint of \(\overline{BD}\), we can conclude that \(\overline{AP}\) is half the length of \(\overline{AB}\). Therefore, \(\overline{AB} = 8\).

Now, since \(\overline{AB} \perp \overline{BC}\) and \(\overline{BD} \perp \overline{DC}\), we can conclude that \(\overline{AB}\) is the height of triangle \(\triangle BCD\). Given that \(AP = 4\) and \(BP = 4\), triangle \(\triangle BCD\) is a rectangle (since opposite sides are equal).

Therefore, the length of \(\overline{DC}\) is also 8.

The perimeter of pentagon \(ABCDP\) is the sum of the lengths of its sides:

\(\text{Perimeter} = \overline{AB} + \overline{BC} + \overline{CD} + \overline{DP} + \overline{PA}\)

We've found that \(\overline{AB} = 8\), \(\overline{BC} = 4\) (since it's half the length of \(\overline{DC}\)), \(\overline{CD} = 8\), \(\overline{DP} = 4\), and \(\overline{PA} = 4\).

So, the perimeter is \(8 + 4 + 8 + 4 + 4 = 28\).