The homework involves calculating the area and width of rectangles and the area of a square given its perimeter. It also includes a table completion task for rectangle dimensions.

Here are the solutions to the problems:

Table Completion:

• Row 1: Length = 3 cm, Width = 6 cm Area = Length $\times$ Width = $3 \text{ cm} \times 6 \text{ cm} = 18 \text{ cm}^2$

• Row 2: Length = 8 m, Width = 15 m Area = Length $\times$ Width = $8 \text{ m} \times 15 \text{ m} = 120 \text{ m}^2$

• Row 3: Width = 9 mm, Area = 36 mm$^2$ Length = Area $\div$ Width = $36 \text{ mm}^2 \div 9 \text{ mm} = 4 \text{ mm}$

• Row 4: Length = 10 km, Area = 15 km$^2$ Width = Area $\div$ Length = $15 \text{ km}^2 \div 10 \text{ km} = 1.5 \text{ km}$

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2. If the length of a rectangle is 36 cm and the area is 720 cm$^2$, what is the width of the rectangle?

Step 1: Write down the given values and the formula for the area of a rectangle.

$$L = 36 \text{ cm}$$ $$A = 720 \text{ cm}^2$$ $$A = L \times W$$

Step 2: Rearrange the formula to solve for the width ($W$).

$$W = \frac{A}{L}$$

Step 3: Substitute the given values into the formula and calculate.

$$W = \frac{720 \text{ cm}^2}{36 \text{ cm}}$$ $$W = 20 \text{ cm}$$

The width of the rectangle is $\boxed{\text{20 cm}}$.

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3. What is the area of a square with a perimeter of 44 m?

Step 1: Write down the given perimeter and the formula for the perimeter of a square.

$$P = 44 \text{ m}$$ $$P = 4s$$

where $s$ is the side length of the square.

Step 2: Solve for the side length ($s$).

$$s = \frac{P}{4}$$ $$s = \frac{44 \text{ m}}{4}$$ $$s = 11 \text{ m}$$

Step 3: Calculate the area of the square using the formula $A = s^2$.

$$A = (11 \text{ m})^2$$ $$A = 121 \text{ m}^2$$

The area of the square is $\boxed{\text{121 m}^2}$.

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4. A square having length 6 m has the same perimeter as a rectangle. Find the area of the rectangle if its length is 4 cm longer than its width.

Assumption: The length of the rectangle is 4 meters longer than its width, to maintain consistent units with the square's dimensions.

Step 1: Calculate the perimeter of the square.
Side length of square $s = 6 \text{ m}$.
Perimeter of square $P_{square} = 4s = 4 \times 6 \text{ m} = 24 \text{ m}$.

Step 2: The perimeter of the rectangle is the same as the square.
Perimeter of rectangle $P_{rect} = 24 \text{ m}$.
Let the length of the rectangle be $L$ and the width be $W$.
We are given $L = W + 4 \text{ m}$.
The formula for the perimeter of a rectangle is $P_{rect} = 2(L+W)$.

Step 3: Substitute the expressions for $P_{rect}$ and $L$ into the perimeter formula and solve for $W$.

$$24 = 2((W+4) + W)$$ $$24 = 2(2W + 4)$$ $$12 = 2W + 4$$ $$12 - 4 = 2W$$ $$8 = 2W$$ $$W = \frac{8}{2}$$ $$W = 4 \text{ m}$$

Step 4: Calculate the length $L$.

$$L = W + 4 \text{ m}$$ $$L = 4 \text{ m} + 4 \text{ m}$$ $$L = 8 \text{ m}$$

Step 5: Calculate the area of the rectangle.

$$A_{rect} = L \times W$$ $$A_{rect} = 8 \text{ m} \times 4 \text{ m}$$ $$A_{rect} = 32 \text{ m}^2$$

The area of the rectangle is $\boxed{\text{32 m}^2}$.

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5. The length of a rectangle is twice its width. If its area is 288 cm$^2$, find the length of the rectangle.

Step 1: Write down the given information.
Let the width of the rectangle be $W$.
The length $L = 2W$.
Area $A = 288 \text{ cm}^2$.

Step 2: Use the area formula $A = L \times W$ and substitute $L = 2W$.

$$288 = (2W) \times W$$ $$288 = 2W^2$$

Step 3: Solve for $W$.

$$W^2 = \frac{288}{2}$$ $$W^2 = 144$$ $$W = \sqrt{144}$$ $$W = 12 \text{ cm}$$

Step 4: Calculate the length $L$.

$$L = 2W$$ $$L = 2 \times 12 \text{ cm}$$ $$L = 24 \text{ cm}$$

The length of the rectangle is $\boxed{\text{24 cm}}$.

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6. The figure shows a hole cut in the middle of a cardboard. Calculate the shaded area.

Step 1: Identify the dimensions of the outer square and the inner square (hole).
Outer square side length = 10 cm.
Inner square side length = 4 cm.

Step 2: Calculate the area of the outer square.
Area of outer square $A_{outer} = \text{side} \times \text{side} = 10 \text{ cm} \times 10 \text{ cm} = 100 \text{ cm}^2$.

Step 3: Calculate the area of the inner square (the hole).
Area of inner square $A_{inner} = \text{side} \times \text{side} = 4 \text{ cm} \times 4 \text{ cm} = 16 \text{ cm}^2$.

Step 4: Calculate the shaded area by subtracting the inner area from the outer area.
Shaded Area = $A_{outer} - A_{inner}$
Shaded Area = $100 \text{ cm}^2 - 16 \text{ cm}^2$
Shaded Area = $84 \text{ cm}^2$
The shaded area is $\boxed{\text{84 cm}^2}$.

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7. PQRS is a square of length 12 cm. A square LMNS of area 25 cm$^2$ is removed from it. What will be the length of LP?

Step 1: Determine the side length of square PQRS.
Side length of PQRS, $s_1 = 12 \text{ cm}$.
So, $PS = 12 \text{ cm}$.

Step 2: Determine the side length of square LMNS.
Area of LMNS, $A_2 = 25 \text{ cm}^2$.
Side length of LMNS, $s_2 = \sqrt{A_2} = \sqrt{25 \text{ cm}^2} = 5 \text{ cm}$.
From the diagram, S is a common vertex, and L is on PS. Thus, $SL$ is a side of the square LMNS.
So, $SL = 5 \text{ cm}$.

Step 3: Calculate the length of LP.
$LP = PS - SL$
$LP = 12 \text{ cm} - 5 \text{ cm}$
$LP = 7 \text{ cm}$
The length of LP is $\boxed{\text{7 cm}}$.

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8. The area of a square is twice the area of a rectangle with dimensions 45 cm and 10 cm. Find the length of the square.

Step 1: Calculate the area of the rectangle.
Length of rectangle $L_{rect} = 45 \text{ cm}$.
Width of rectangle $W_{rect} = 10 \text{ cm}$.
Area of rectangle $A_{rect} = L_{rect} \times W_{rect} = 45 \text{ cm} \times 10 \text{ cm} = 450 \text{ cm}^2$.

Step 2: Calculate the area of the square.
Area of square $A_{square} = 2 \times A_{rect}$
$A_{square} = 2 \times 450 \text{ cm}^2 = 900 \text{ cm}^2$.

Step 3: Find the length of the square's side ($s$).
For a square, $A_{square} = s^2$.

$$s = \sqrt{A_{square}}$$ $$s = \sqrt{900 \text{ cm}^2}$$ $$s = 30 \text{ cm}$$

The length of the square is $\boxed{\text{30 cm}}$.

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9. The width of a rectangle is 4 cm less than its length and its perimeter is 48 cm. Find its area.

Step 1: Define the length and width in terms of a single variable.
Let the length of the rectangle be $L$.
The width $W = L - 4 \text{ cm}$.

Step 2: Use the given perimeter and the perimeter formula.
Perimeter $P = 48 \text{ cm}$.
$P = 2(L+W)$.

Step 3: Substitute $W = L-4$ into the perimeter formula and solve for $L$.

$$48 = 2(L + (L-4))$$ $$48 = 2(2L - 4)$$ $$24 = 2L - 4$$ $$24 + 4 = 2L$$ $$28 = 2L$$ $$L = \frac{28}{2}$$ $$L = 14 \text{ cm}$$

Step 4: Calculate the width $W$.

$$W = L - 4 \text{ cm}$$ $$$$