Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

The base and height of the right-angled triangle are given as:

• Base: $x$ cm
• Height: $2x$ cm

The area of the triangle $A_{triangle}$ can be expressed as: $A_{triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times x \times 2x = x^2$

To find the set of possible values for $x$, we need to set up the inequality based on the problem statement: $A_{rectangle} > A_{triangle}$ This translates to: $3x^2 - 3x - 2 > x^2$

Rearranging the inequality gives: $3x^2 - 3x - 2 - x^2 > 0$ Which simplifies to: $2x^2 - 3x - 2 > 0$

To solve the inequality $2x^2 - 3x - 2 > 0$, we first find the roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 2$, $b = -3$, and $c = -2$: $x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$ This gives: $x = 2 \quad \text{(approximately)} \quad \text{or} \quad x = -\frac{1}{2}$ Hence, the critical points are $x = 2$ and $x = -\frac{1}{2}$.

To analyze the intervals:

• Test points in the intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2}, 2)$, and $(2, +\infty)$ to find where the inequality holds true.

After testing these intervals, we find that the inequality holds for: $x > 2$.