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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.

Let's solve the problem step by step:

### (a) Form an equation and simplify to \( 2x^2 + 5x - 52 = 0 \)

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

\[
A = \frac{1}{2} \times (PQ + RS) \times \text{height}
\]

From the diagram, we know:
- \( PQ = x \, \text{cm} \)
- \( RS = 3x - 2 \, \text{cm} \)
- The height is \( PS = x + 3 \, \text{cm} \)

The area is given as 49 cm². Therefore, we substitute these values into the area formula:

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

Simplify the expression inside the parentheses:

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

Multiply both sides by 2 to eliminate the fraction:

\[
98 = (4x - 2)(x + 3)
\]

Now expand the right-hand side:

\[
98 = 4x(x + 3) - 2(x + 3)
\]

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

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

Now move all terms to the left-hand side:

\[
4x^2 + 10x - 6 - 98 = 0
\]

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

Finally, divide the entire equation by 2:

\[
2x^2 + 5x - 52 = 0
\]

This is the required equation.

---

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

To solve this quadratic equation, we can use the quadratic formula:

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

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

Now, calculate the discriminant:

\[
\Delta = b^2 - 4ac = 5^2 - 4(2)(-52)
\]

\[
\Delta = 25 + 416 = 441
\]

Since the discriminant is positive, we have two real solutions. Now apply the quadratic formula:

\[
x = \frac{-5 \pm \sqrt{441}}{2(2)}
\]

\[
x = \frac{-5 \pm 21}{4}
\]

This gives two possible values for \( x \):

\[
x = \frac{-5 + 21}{4} = \frac{16}{4} = 4
\]

or

\[
x = \frac{-5 - 21}{4} = \frac{-26}{4} = -6.5
\]

Since \( x \) represents a length, we discard the negative solution, so \( x = 4 \).

---

### (c) Find the length of RS

Now that we know \( x = 4 \), we can substitute this into the expression for \( RS \):

\[
RS = 3x - 2 = 3(4) - 2 = 12 - 2 = 10 \, \text{cm}
\]

Thus, the length of \( RS \) is 10 cm.

---

Would you like further details or have any questions?

---

### Related Questions:
1. How would the area formula change if \( PQ \) and \( RS \) were not parallel?
2. What is the significance of the discriminant in the quadratic formula?
3. Can you solve the equation using factoring instead of the quadratic formula?
4. How can we verify that \( x = 4 \) is the correct solution using a different method?
5. What happens to the dimensions of the trapezium if the area is doubled?

### Tip:
When solving quadratic equations, always check whether the discriminant is positive, zero, or negative to determine the nature of the roots (real or complex).

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 quadratic equation derived from the area of a trapezium PQRS. This problem involves algebra, quadratic equations, and geometry concepts. Includes step-by-step solutions to form the equation 2x² + 5x - 52 = 0, solve it, and determine the length of RS. Suitable for Grades 9-10.

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