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The diagram shows trapezium PQRS. PQ is parallel to RS and angle PSR = 90°. PQ = x cm, PS = (x + 3) cm and RS = (3x - 2) cm.
(a) Given that the area of the trapezium is 49 cm?, form an equation in
x and show that it simplifies to 2x* + 5x - 52 = 0.
(b) Solve the equation 2x2 + 5x - 52 = 0.
(c) Hence, find the length of RS.

### Step-by-step Solution

The trapezium \(PQRS\) has the following dimensions:
- \(PQ = x \, \text{cm}\)
- \(PS = x + 3 \, \text{cm}\)
- \(RS = 3x - 2 \, \text{cm}\)
- \(\angle PSR = 90^\circ\), making \(PS \perp RS\).

#### Part (a): Form an equation for the area of the trapezium

The area \(A\) of a trapezium is given by the formula:

\[
A = \frac{1}{2} \times (a + b) \times h
\]

Where:
- \(a\) and \(b\) are the lengths of the parallel sides (\(PQ\) and \(RS\)),
- \(h\) is the perpendicular height (which is \(PS\) in this case).

Substituting the given values:
\[
PQ = x \, \text{cm}, \quad RS = (3x - 2) \, \text{cm}, \quad PS = (x + 3) \, \text{cm}
\]

The area of the trapezium is given as \(49 \, \text{cm}^2\). Therefore, substituting into the area formula:

\[
49 = \frac{1}{2} \times (x + (3x - 2)) \times (x + 3)
\]

Simplifying:
\[
49 = \frac{1}{2} \times (4x - 2) \times (x + 3)
\]

Multiply both sides by 2 to eliminate the fraction:
\[
98 = (4x - 2)(x + 3)
\]

Now, expand the right-hand side:
\[
98 = 4x(x + 3) - 2(x + 3)
\]

\[
98 = 4x^2 + 12x - 2x - 6
\]

\[
98 = 4x^2 + 10x - 6
\]

Subtract 98 from both sides:
\[
0 = 4x^2 + 10x - 6 - 98
\]

\[
0 = 4x^2 + 10x - 104
\]

Now divide the entire equation by 2 to simplify:
\[
0 = 2x^2 + 5x - 52
\]

This is the required equation.

#### Part (b): Solve the quadratic equation \(2x^2 + 5x - 52 = 0\)

We will use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For the equation \(2x^2 + 5x - 52 = 0\), we have:
- \(a = 2\),
- \(b = 5\),
- \(c = -52\).

First, calculate the discriminant:
\[
\Delta = b^2 - 4ac = 5^2 - 4(2)(-52)
\]
\[
\Delta = 25 + 416 = 441
\]

Now, substitute into the quadratic formula:
\[
x = \frac{-5 \pm \sqrt{441}}{2(2)}
\]
\[
x = \frac{-5 \pm 21}{4}
\]

Thus, the two possible values for \(x\) are:
\[
x_1 = \frac{-5 + 21}{4} = \frac{16}{4} = 4
\]
\[
x_2 = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5
\]

Since \(x\) represents a length, it must be positive. Therefore, \(x = 4\).

#### Part (c): Find the length of \(RS\)

Now that we know \(x = 4\), we can find the length of \(RS\):
\[
RS = 3x - 2 = 3(4) - 2 = 12 - 2 = 10 \, \text{cm}
\]

### Final Answers:
- (a) The equation simplifies to \(2x^2 + 5x - 52 = 0\).
- (b) The solution to the equation is \(x = 4\).
- (c) The length of \(RS\) is \(10 \, \text{cm}\).

Let me know if you need any additional details or have questions!

---

**Questions for deeper understanding:**
1. How would the equation change if the area of the trapezium were different?
2. What are the properties of trapeziums that help in solving such problems?
3. Why did we discard the negative value of \(x\) in the solution?
4. How would you calculate the perimeter of trapezium \(PQRS\) using the values obtained?
5. What would be the next steps if the angle \(\angle PSR\) was different?

**Tip:** When using the quadratic formula, always check whether both roots are applicable based on the context of the problem (e.g., lengths must be positive).

The diagram shows trapezium PQRS. PQ is parallel to RS and angle PSR = 90°. PQ = x cm, PS = (x + 3) cm and RS = (3x - 2) cm. (a) Given that the area of the trapezium is 49 cm², form an equation in x and show that it simplifies to 2x² + 5x - 52 = 0. (b) Solve the equation 2x² + 5x - 52 = 0. (c) Hence, find the length of RS.

Learn how to solve a geometry problem involving a trapezium PQRS where PQ is parallel to RS and angle PSR = 90°. The problem requires forming a quadratic equation and solving for x using algebraic techniques. Suitable for Grades 8-10 students.

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