Question

Square root or cube root to find perimeter and area
7√32 length nd width 3√2

Answer 1

To find the perimeter and area using a length of \( 7\sqrt{32} \) and a width of \( 3\sqrt{2} \), we can calculate each as follows:

### Area
The area \( A \) of a rectangle is given by the formula:
\[
A = \text{length} \times \text{width}
\]
Substituting the provided values:
\[
A = 7\sqrt{32} \times 3\sqrt{2}
\]

First, calculate \( \sqrt{32} \):
\[
\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}
\]

Now substitute \( \sqrt{32} \) back into the area calculation:
\[
A = 7(4\sqrt{2}) \times 3\sqrt{2}
\]
\[
A = 28\sqrt{2} \times 3\sqrt{2}
\]
\[
A = 84 \cdot (\sqrt{2} \times \sqrt{2}) = 84 \cdot 2 = 168
\]

So, the area is:
\[
\text{Area} = 168
\]

### Perimeter
The perimeter \( P \) of a rectangle is given by the formula:
\[
P = 2 \times (\text{length} + \text{width})
\]
Substituting the provided values:
\[
P = 2 \times (7\sqrt{32} + 3\sqrt{2})
\]
Again, substitute \( \sqrt{32} = 4\sqrt{2} \):
\[
P = 2 \times (7(4\sqrt{2}) + 3\sqrt{2}) = 2 \times (28\sqrt{2} + 3\sqrt{2}) = 2 \times (31\sqrt{2})
\]
\[
P = 62\sqrt{2}
\]

So, the perimeter is:
\[
\text{Perimeter} = 62\sqrt{2}
\]

### Summary
- Area = \( 168 \)
- Perimeter = \( 62\sqrt{2} \)