https://file.aiquickdraw.com/mathbot/file/1728876932344512x8jm6.jpg

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.
P
(x + 3) cm
X Cr
(b) Solve the equation 2x2 + 5x - 52 = 0.
(c) Hence, find the length of RS.

Let's break down the problem step by step based on the image provided. The shape is a trapezium \( PQRS \), with \( PQ \parallel RS \) and \( \angle PSR = 90^\circ \). We are given the following dimensions:

- \( PQ = x \) cm (parallel side),
- \( PS = (x + 3) \) cm (height),
- \( RS = (3x - 2) \) cm (parallel side).

We are asked to:

### (a) Form an equation in \( x \) for the area of the trapezium and simplify it.

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

\[
A = \frac{1}{2} \times \text{sum of parallel sides} \times \text{height}
\]

Here, the parallel sides are \( PQ = x \) and \( RS = (3x - 2) \), and the height is \( PS = (x + 3) \). The area is given as 49 cm². So, substituting into the formula:

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

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 trapezium area problem involving a quadratic equation. Given the area of the trapezium, form the equation and simplify it to 2x² + 5x - 52 = 0. Includes solving the quadratic equation and finding the length of RS. Suitable for students in Grades 8-10.

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