Worksheet with polynomial word problems and geometric diagrams.

Let's solve each problem step by step.

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Problem 1:

One side of a square is \( (2x - 5) \). What is the area of the square?

Solution:
The area of a square is given by the formula:
\[
\text{Area} = (\text{side})^2
\]
Here, the side of the square is \( 2x - 5 \). Therefore, the area is:
\[
\text{Area} = (2x - 5)^2
\]
Expanding the square:
\[
(2x - 5)^2 = (2x)^2 - 2 \cdot 2x \cdot 5 + 5^2 = 4x^2 - 20x + 25
\]
So, the area of the square is:
\[
\boxed{4x^2 - 20x + 25}
\]

---

Problem 2:

If the perimeter of a rectangle is \( (32x + 20y) \) and its width is \( (2x + 2y) \), what is the length of the rectangle?

Solution:
The perimeter of a rectangle is given by the formula:
\[
\text{Perimeter} = 2 \times (\text{length} + \text{width})
\]
Let the length of the rectangle be \( L \). The width is given as \( 2x + 2y \). Substituting the given perimeter:
\[
32x + 20y = 2 \times (L + 2x + 2y)
\]
Divide both sides by 2:
\[
16x + 10y = L + 2x + 2y
\]
Isolate \( L \) by subtracting \( 2x + 2y \) from both sides:
\[
L = 16x + 10y - 2x - 2y = 14x + 8y
\]
So, the length of the rectangle is:
\[
\boxed{14x + 8y}
\]

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Problem 3:

The side of a cube measures \( (x + 3) \). What is the volume of the cube?

Solution:
The volume of a cube is given by the formula:
\[
\text{Volume} = (\text{side})^3
\]
Here, the side of the cube is \( x + 3 \). Therefore, the volume is:
\[
\text{Volume} = (x + 3)^3
\]
Expanding the cube:
\[
(x + 3)^3 = (x + 3)(x + 3)(x + 3)
\]
First, expand \( (x + 3)(x + 3) \):
\[
(x + 3)(x + 3) = x^2 + 6x + 9
\]
Now, multiply by \( x + 3 \):
\[
(x^2 + 6x + 9)(x + 3) = x^3 + 3x^2 + 6x^2 + 18x + 9x + 27 = x^3 + 9x^2 + 27x + 27
\]
So, the volume of the cube is:
\[
\boxed{x^3 + 9x^2 + 27x + 27}
\]

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Problem 4:

The measures of two sides of a triangle are \( (x + 4) \) and \( (4x - 7) \). If the perimeter of the triangle is \( (19x + 8) \), what is the measure of the third side?

Solution:
The perimeter of a triangle is the sum of the lengths of its three sides. Let the third side be \( S \). The perimeter is given as:
\[
(x + 4) + (4x - 7) + S = 19x + 8
\]
Simplify the left side:
\[
x + 4 + 4x - 7 + S = 19x + 8
\]
\[
5x - 3 + S = 19x + 8
\]
Isolate \( S \) by subtracting \( 5x - 3 \) from both sides:
\[
S = 19x + 8 - 5x + 3 = 14x + 11
\]
So, the measure of the third side is:
\[
\boxed{14x + 11}
\]

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Problem 5:

Find the area of the shaded region.

Solution:
The shaded region is the area of the larger rectangle minus the area of the smaller square.

1. Area of the larger rectangle:
The dimensions of the larger rectangle are \( 5x + 3 \) (width) and \( 2x - 5 \) (height). The area is:
\[
\text{Area of the rectangle} = (5x + 3)(2x - 5)
\]
Expand the product:
\[
(5x + 3)(2x - 5) = 5x \cdot 2x + 5x \cdot (-5) + 3 \cdot 2x + 3 \cdot (-5) = 10x^2 - 25x + 6x - 15 = 10x^2 - 19x - 15
\]

2. Area of the smaller square:
The side of the smaller square is 3. The area is:
\[
\text{Area of the square} = 3^2 = 9
\]

3. Area of the shaded region:
Subtract the area of the square from the area of the rectangle:
\[
\text{Shaded area} = (10x^2 - 19x - 15) - 9 = 10x^2 - 19x - 24
\]
So, the area of the shaded region is:
\[
\boxed{10x^2 - 19x - 24}
\]

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Problem 6:

Write an expression for the perimeter of the triangle.

Solution:
The perimeter of a triangle is the sum of the lengths of its three sides. The sides are given as \( 5x - 7 \), \( 5x - 7 \), and \( 6x \). The perimeter is:
\[
\text{Perimeter} = (5x - 7) + (5x - 7) + 6x = 5x - 7 + 5x - 7 + 6x = 16x - 14
\]
So, the perimeter is:
\[
\boxed{16x - 14}
\]

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Problem 7:

Write an expression for the perimeter of the trapezoid.

Solution:
The perimeter of a trapezoid is the sum of the lengths of its four sides. The sides are given as \( 19x + 12 \), \( 5x - 3 \), \( 12x - 8 \), and the fourth side (which is not labeled but can be assumed to be part of the trapezoid). However, since the fourth side is not explicitly given, we assume it is part of the trapezoid and include it in the perimeter. The perimeter is:
\[
\text{Perimeter} = (19x + 12) + (5x - 3) + (12x - 8) + \text{(fourth side)}
\]
Without the fourth side, the expression for the perimeter of the three given sides is:
\[
(19x + 12) + (5x - 3) + (12x - 8) = 19x + 5x + 12x + 12 - 3 - 8 = 36x + 1
\]
So, the perimeter is:
\[
\boxed{36x + 1}
\]

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Problem 8:

The height of a triangle is \( (6x - 8) \) and its base is \( (2x + 6) \). What is the area of the triangle?

Solution:
The area of a triangle is given by the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Here, the base is \( 2x + 6 \) and the height is \( 6x - 8 \). The area is:
\[
\text{Area} = \frac{1}{2} \times (2x + 6) \times (6x - 8)
\]
First, expand the product \( (2x + 6)(6x - 8) \):
\[
(2x + 6)(6x - 8) = 2x \cdot 6x + 2x \cdot (-8) + 6 \cdot 6x + 6 \cdot (-8) = 12x^2 - 16x + 36x - 48 = 12x^2 + 20x - 48
\]
Now, multiply by \( \frac{1}{2} \):
\[
\text{Area} = \frac{1}{2} \times (12x^2 + 20x - 48) = 6x^2 + 10x - 24
\]
So, the area of the triangle is:
\[
\boxed{6x^2 + 10x - 24}
\]

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Final Answers:

1. \(\boxed{4x^2 - 20x + 25}\)
2. \(\boxed{14x + 8y}\)
3. \(\boxed{x^3 + 9x^2 + 27x + 27}\)
4. \(\boxed{14x + 11}\)
5. \(\boxed{10x^2 - 19x - 24}\)
6. \(\boxed{16x - 14}\)
7. \(\boxed{36x + 1}\)
8. \(\boxed{6x^2 + 10x - 24}\)