9.7. Let square $\mathrm{ABCD}$ have a side length of $3 \mathrm{~cm}$. Points $\mathrm{M}, \mathrm{N}$, and $\mathrm{P}$ are located on sides (AB), (BC), and (CD), respectively, such that quadrilateral AMNP is a trapezoid with bases MN and AP, having perpendicular diagonals and $B N=1 \mathrm{~cm}$. Calculate the perimeter of trapezoid AMNP.
Solution. Since $B N \| A D$ and $M N \| A P$, then angles BNM and DAP are congruent, which implies the similarity of triangles BNM and DAP. Since $B N=1 \mathrm{~cm}$, and $A D=3 \mathrm{~cm}$, it follows that $A P=3 \cdot M N$. If $M P \cap A N=\{O\}$, then from the congruence of angles OMN and OPA, the similarity of triangles OMN and OPA follows. Thus, we obtain the relationships $O P=3 \cdot O M, O A=3 O N$. From

![](https://cdn.mathpix.com/cropped/2024_06_06_913029280265c0e289a3g-1.jpg?height=583&width=610&top_left_y=1969&top_left_x=1365)
triangle ABN, using the Pythagorean theorem, we obtain the length

of segment AN: $A N^{2}=B N^{2}+B A^{2}=1^{2}+3^{2}=10$, from which

$$
A N=\sqrt{10} \mathrm{~cm}, O N=\frac{\sqrt{10}}{4} \mathrm{~cm}, O A=\frac{3 \sqrt{10}}{4} \mathrm{~cm}
$$

Since angles OAM and BAN are congruent, it follows that triangles OAM and BAN are similar, from which we obtain the relationships:

$$
\frac{O A}{B A}=\frac{O M}{B N}=\frac{A M}{A N} \Leftrightarrow \frac{\sqrt{10}}{4}=\frac{O M}{1}=\frac{A M}{\sqrt{10}} \Leftrightarrow O M=\frac{\sqrt{10}}{4} \mathrm{~cm}, A M=\frac{5}{2} \mathrm{~cm}
$$

Since $O M=O N=\frac{\sqrt{10}}{4} \mathrm{~cm}$, it follows that triangles OMN and OPA are right isosceles, which shows that trapezoid AMNP is isosceles with $A M=N P=\frac{5}{2} \mathrm{~cm}$. Since $B M=A B-A M=3-\frac{5}{2}=$ $\frac{1}{2}(\mathrm{~cm})$, from triangle MBN, using the Pythagorean theorem, we obtain the lengths of segments MN and PA:

$$
M N^{2}=B M^{2}+B N^{2}=\frac{1}{4}+1=\frac{5}{4} \Rightarrow M N=\frac{\sqrt{5}}{2} \mathrm{~cm}, \quad P A=\frac{3 \sqrt{5}}{2} \mathrm{~cm}
$$

Then the perimeter of trapezoid AMNP will be equal to $P(\operatorname{tr} . A M N P)=(5+2 \sqrt{5}) \mathrm{cm}$.