Let's break the problem into parts and solve each step-by-step.

Given:

1. Quadrilateral $ABCD$.
2. $\angle ABC = \angle CAD = 90^\circ$.
3. $|AB| = 4 \, \text{cm}, |BC| = 3 \, \text{cm}, |AD| = 12 \, \text{cm}$.

We need to find:

1. $|AC|$ and $|CD|$.
2. Perimeter of quadrilateral $ABCD$.
3. Area of triangle $ACD$.

Solution:

1. Finding $|AC|$ and $|CD|$:

• Finding $|AC|$: Using the Pythagorean theorem in $\triangle ABC$: $|AC|^2 = |AB|^2 + |BC|^2$ $|AC|^2 = 4^2 + 3^2 = 16 + 9 = 25$ $|AC| = \sqrt{25} = 5 \, \text{cm}.$

• Finding $|CD|$: In $\triangle ACD$, $AC = 5 \, \text{cm}$ (calculated above), and $AD = 12 \, \text{cm}$. Again using the Pythagorean theorem: $|CD|^2 = |AD|^2 - |AC|^2$ $|CD|^2 = 12^2 - 5^2 = 144 - 25 = 119$ $|CD| = \sqrt{119} \approx 10.91 \, \text{cm}.$

2. Perimeter of $ABCD$:

The perimeter is the sum of all side lengths: $\text{Perimeter} = |AB| + |BC| + |CD| + |DA|$ $\text{Perimeter} = 4 + 3 + 10.91 + 12 = 29.91 \, \text{cm}.$

3. Area of triangle $ACD$:

The area of $\triangle ACD$ is given by: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}.$ Here, $AC$ is the base, and $AD$ is the height: $\text{Area} = \frac{1}{2} \times 5 \times 12 = 30 \, \text{cm}^2.$

Final Answers:

1. $|AC| = 5 \, \text{cm}, |CD| \approx 10.91 \, \text{cm}$.
2. Perimeter of $ABCD \approx 29.91 \, \text{cm}$.
3. Area of $\triangle ACD = 30 \, \text{cm}^2$.

Do you want a detailed explanation for any part? Let me know!

Here are 5 related questions:

1. How do you find the area of a quadrilateral if it is not rectangular?
2. Can we calculate the diagonal $BD$ of $ABCD$?
3. What happens to the perimeter if $AD$ increases?
4. How do the Pythagorean theorem and trigonometry relate?
5. Could we use trigonometric functions to find angles in the diagram?

Tip: Always double-check given dimensions before applying geometric theorems.