To solve for $x$ in this trapezium problem, we will use the fact that in any trapezium where $AB \parallel CD$, if the diagonals are drawn and intersect at point $O$, the following property holds:

$\frac{OA}{OB} = \frac{OC}{OD}$

Given:

• $OA = x + 5$
• $OB = x - 7$
• $OC = x + 3$
• $OD = x - 2$

Using the property, we can set up the equation:

$\frac{OA}{OB} = \frac{OC}{OD}$

Substituting the given values:

$\frac{x + 5}{x - 7} = \frac{x + 3}{x - 2}$

Now, we will solve this equation step-by-step.

1. Cross-multiply:

$(x + 5)(x - 2) = (x - 7)(x + 3)$

2. Expand both sides:

$(x + 5)(x - 2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10$ $(x - 7)(x + 3) = x^2 + 3x - 7x - 21 = x^2 - 4x - 21$

3. Set the expanded forms equal to each other:

$x^2 + 3x - 10 = x^2 - 4x - 21$

4. Simplify by subtracting $x^2$ from both sides:

$3x - 10 = -4x - 21$

5. Add $4x$ to both sides:

$7x - 10 = -21$

6. Add 10 to both sides:

$7x = -11$

7. Divide by 7:

$x = -\frac{11}{7}$

Thus, the value of $x$ is $-\frac{11}{7}$.

Do you want more details, or do you have any questions?

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Here are 5 related questions:

1. What is the formula for the area of a trapezium?
2. How do the properties of parallel sides in a trapezium affect its diagonals?
3. Can the diagonals of a trapezium be equal? Under what conditions?
4. What are the different types of trapeziums, and how do their properties differ?
5. How do you derive the equation of a line parallel to another in geometry?

Tip: In a trapezium, the diagonals generally do not bisect each other, but they divide each other proportionally when the sides are parallel.