https://r2.erweima.ai/mathbot/image/1726113578647orycwlaj.jpg

Let's go through the questions in the image one by one and solve them.

### 4. Obtain the perimeter
For the figures in problems 63 and 64, we need to find expressions for the perimeter of each shape by adding up the lengths of all sides.

**Problem 63:**
The figure has four sides, with the following side lengths:
- Top side: \( 4x^2 - 5x \)
- Bottom side: \( 6x + x \) (which simplifies to \( 7x \))
- Left side: \( 2x^2 + 6x \)
- Right side: \( 3x + 7 \)

The perimeter \( P \) is the sum of all these sides:
\[
P = (2x^2 + 6x) + (4x^2 - 5x) + (7x) + (3x + 7)
\]
Simplifying:
\[
P = 2x^2 + 6x + 4x^2 - 5x + 7x + 3x + 7
\]
\[
P = 6x^2 + 11x + 7
\]

**Problem 64:**
The figure is a quadrilateral with four sides:
- Top side: \( 5x - 2 \)
- Bottom side: \( 2x - 6 \)
- Left side: \( 3x - 1 \)
- Right side: \( 2x - 6 \)

The perimeter \( P \) is the sum of all these sides:
\[
P = (5x - 2) + (2x - 6) + (3x - 1) + (2x - 6)
\]
Simplifying:
\[
P = 5x - 2 + 2x - 6 + 3x - 1 + 2x - 6
\]
\[
P = 12x - 15
\]

### 5. Obtain an expression for the area of the following figures

**Problem 99 (Square):**
For a square, the area \( A \) is given by:
\[
A = \text{side}^2
\]
The side length is \( x - 2 \), so the area is:
\[
A = (x - 2)^2
\]
Expanding:
\[
A = x^2 - 4x + 4
\]

**Problem 100 (Square):**
The side length is \( x + 3 \), so the area is:
\[
A = (x + 3)^2
\]
Expanding:
\[
A = x^2 + 6x + 9
\]

**Problem 101 (Rectangle):**
The area of a rectangle is given by:
\[
A = \text{length} \times \text{width}
\]
The dimensions are \( x - 2 \) and \( x + 2 \), so the area is:
\[
A = (x - 2)(x + 2)
\]
This is a difference of squares:
\[
A = x^2 - 4
\]

**Problem 102 (Rectangle):**
The dimensions are \( 2x - 3 \) and \( 2x + 3 \), so the area is:
\[
A = (2x - 3)(2x + 3)
\]
This is also a difference of squares:
\[
A = 4x^2 - 9
\]

**Problem 103 (Triangle):**
The area of a triangle is given by:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
The base is \( 2x - 6 \) and the height is \( x + 3 \), so the area is:
\[
A = \frac{1}{2} \times (2x - 6) \times (x + 3)
\]
Simplifying:
\[
A = \frac{1}{2} \times [2x^2 + 6x - 18]
\]
\[
A = x^2 + 3x - 9
\]

**Problem 104 (Triangle):**
The base is \( x - 1 \) and the height is \( 4x \), so the area is:
\[
A = \frac{1}{2} \times (x - 1) \times 4x
\]
Simplifying:
\[
A = 2x(x - 1) = 2x^2 - 2x
\]

### 6. Obtain an expression for the volume of the following figures

**Problem 105 (Rectangular prism):**
The dimensions are \( 3x + 10 \), \( 3x \), and the depth is not shown, so the volume is simply:
\[
V = \text{length} \times \text{width} \times \text{height}
\]
Assuming the depth \( d \) is given.

**Problem 106 (Rectangular prism):**
Same process, dimensions not fully given.

For more precise details, I recommend labeling the images clearer for problem 6. 

Would you like to discuss more? Do these results match up? Or tips?

Solve the problems related to perimeter, area, and volume of shapes as shown in the image.

This guide covers solving perimeter, area, and volume problems for various shapes including squares, rectangles, triangles, and cubes. Learn how to apply algebraic expressions to calculate the perimeter of polygons and find areas and volumes of different figures.

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