Given:

• Quadrilateral ABCD with:
  • $AB = 4 \, \text{cm}$
  • $BC = 3 \, \text{cm}$
  • $AD = 12 \, \text{cm}$
• $\angle ABC = \angle CAD = 90^\circ$.

We are tasked with:

1. Finding the lengths $AC$ and $CD$,
2. Calculating the perimeter of quadrilateral $ABCD$,
3. Calculating the area of triangle $ACD$.

Solution:

1. Finding $AC$ and $CD$:

(a) $AC$:

From triangle $ABC$, use Pythagoras' theorem: $AC^2 = AB^2 + BC^2$ $AC^2 = 4^2 + 3^2 = 16 + 9 = 25$ $AC = \sqrt{25} = 5 \, \text{cm}.$

(b) $CD$:

From triangle $ACD$, again use Pythagoras' theorem. The hypotenuse is $AD = 12 \, \text{cm}$, and one leg is $AC = 5 \, \text{cm}$: $CD^2 = AD^2 - AC^2$ $CD^2 = 12^2 - 5^2 = 144 - 25 = 119$ $CD = \sqrt{119} \approx 10.91 \, \text{cm}.$

2. Perimeter of $ABCD$:

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

3. Area of triangle $ACD$:

The area of triangle $ACD$ can be found using the formula: $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}.$ Here:

• Base $AC = 5 \, \text{cm}$,
• Height $CD \approx 10.91 \, \text{cm}$, $\text{Area} = \frac{1}{2} \times 5 \times 10.91 = 27.275 \, \text{cm}^2 \, \text{(approximately)}.$

Final Answers:

1. $AC = 5 \, \text{cm}, \, CD \approx 10.91 \, \text{cm}$,
2. Perimeter $\approx 29.91 \, \text{cm}$,
3. Area of $\triangle ACD \approx 27.28 \, \text{cm}^2$.

Would you like additional steps or further clarifications?

Related Questions:

1. What are the properties of quadrilaterals with two right angles like $ABCD$?
2. How does the Pythagorean theorem apply to composite shapes?
3. Can you derive $CD$ using trigonometry instead of Pythagoras' theorem?
4. How does the area formula for triangles generalize to non-right-angled triangles?
5. What is the significance of approximate calculations in geometry?

Tip:

Always double-check units and whether results make sense geometrically to avoid errors!