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Let's solve the two parts of this problem step by step.

### Given:
- Width of the rectangle = \( \sqrt{12} \) cm
- Height of the rectangle = \( \sqrt{27} \) cm

### Part a) Calculate the area of the rectangle.

The area \( A \) of a rectangle is calculated using the formula:
\[
A = \text{Width} \times \text{Height}
\]
Substitute the given values:
\[
A = \sqrt{12} \times \sqrt{27}
\]
Using the property \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \):
\[
A = \sqrt{12 \times 27} = \sqrt{324}
\]
Since \( \sqrt{324} = 18 \), the area is:
\[
A = 18 \, \text{cm}^2
\]

### Part b) Calculate the perimeter of the rectangle.

The perimeter \( P \) of a rectangle is given by:
\[
P = 2 \times (\text{Width} + \text{Height})
\]
Substitute the given values:
\[
P = 2 \times (\sqrt{12} + \sqrt{27})
\]

We simplify the surds:
\[
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}, \quad \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}
\]

Now, add these values:
\[
\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}
\]

Finally, multiply by 2 to get the perimeter:
\[
P = 2 \times 5\sqrt{3} = 10\sqrt{3} \, \text{cm}
\]

### Final Answers:
- Area: \( 18 \, \text{cm}^2 \)
- Perimeter: \( 10\sqrt{3} \, \text{cm} \)

Would you like further details or have any questions?

Here are 5 related questions for further practice:
1. What is the length of the diagonal of the rectangle?
2. How would the perimeter change if the width and height were doubled?
3. What is the simplified value of \( \sqrt{3} \) if you were to approximate the perimeter numerically?
4. Can you calculate the area of a square with a side length of \( \sqrt{50} \) cm?
5. How would you calculate the area and perimeter if the sides were given as decimals instead of surds?

**Tip:** When multiplying or simplifying surds, look for perfect square factors to make the process easier!

a) Calculate the area, in cm^2, of the rectangle.
b) Calculate the perimeter, in cm, of the rectangle.
Given: Width = √12 cm, Height = √27 cm

Learn how to calculate the area and perimeter of a rectangle with surds √12 cm and √27 cm as its width and height, respectively. This problem involves simplifying surds and applying formulas for area and perimeter. Suitable for students in Grades 8-10.

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