The problem involves a right trapezoid $ABCD$ with one right angle at $A$. Here's a detailed translation and breakdown of the question:

Given:

• $AD$ is the height (one of the bases) of the right trapezoid and $AD = 12 \, \text{cm}$.
• $AB = 5 \, \text{cm}$ (the shorter parallel side).
• $\angle D = 45^\circ$.

Find:

• The lengths of the vectors:
  • $\vec{BD}$
  • $\vec{CD}$
  • $\vec{AC}$.

Solution:

1. Understanding the Trapezoid Geometry:

   • Since $AD$ is perpendicular to $AB$, $\triangle ABD$ is a right triangle.
   • The angle $\angle D = 45^\circ$ means $\triangle ABD$ is a 45°-45°-90° triangle.
   • In a 45°-45°-90° triangle, the legs are equal, so $AD = BD = 12 \, \text{cm}$.

2. Finding $CD$:

   • $AB \parallel CD$, so $CD = AB + BD$.
   • Substituting values: $CD = 5 + 12 = 17 \, \text{cm}.$

3. Finding the Lengths of Vectors:

   • $\vec{BD}$: The length of $\vec{BD}$ is simply the distance $BD = 12 \, \text{cm}$.

   • $\vec{CD}$: The length of $\vec{CD}$ is the distance $CD = 17 \, \text{cm}$.

   • $\vec{AC}$: In $\triangle ACD$, we can apply the Pythagorean theorem to find $AC$: $AC = \sqrt{AD^2 + CD^2}.$ Substituting $AD = 12$ and $CD = 17$: $AC = \sqrt{12^2 + 17^2} = \sqrt{144 + 289} = \sqrt{433}.$ Thus, $AC \approx 20.81 \, \text{cm}$.

Final Answer:

• $|\vec{BD}| = 12 \, \text{cm}$
• $|\vec{CD}| = 17 \, \text{cm}$
• $|\vec{AC}| \approx 20.81 \, \text{cm}$

Would you like me to provide more details, a diagram, or additional explanations?

Related Questions:

1. How do you calculate the angles in a 45°-45°-90° triangle?
2. What is the geometric meaning of vector magnitudes in this context?
3. Can vector directions also be determined for this problem?
4. How do parallel lines affect the dimensions of a trapezoid?
5. Why does the Pythagorean theorem work for finding $AC$ in this scenario?

Tip:

In a 45°-45°-90° triangle, the hypotenuse is always $\sqrt{2}$ times the length of one leg.