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

The triangle has a base of $x$ and a height of $(2x)$. The area of the triangle (A_triangle) can thus be represented as:

$A_{triangle} = \frac{1}{2} \times base \times height = \frac{1}{2} \times x \times (2x) = x^2$.

According to the problem, the area of the rectangle must be greater than the area of the triangle. Hence, we set up the inequality:

$3x^2 - 2x - 3 > x^2$.

We simplify the inequality:

$3x^2 - 2x - 3 - x^2 > 0$ $2x^2 - 2x - 3 > 0$ $x^2 - x - \frac{3}{2} > 0$

Next, we need to find the roots of the equation:

$x^2 - x - \frac{3}{2} = 0$ Using the quadratic formula, where $a=1$, $b=-1$, and $c=-\frac{3}{2}$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{1 + 6}}{2} = \frac{1 \pm \sqrt{7}}{2}$

This gives the critical points:

$x = \frac{1 + \sqrt{7}}{2}$ and $x = \frac{1 - \sqrt{7}}{2}$

We find these critical points and test intervals to determine in which intervals the inequality holds true.

Based on the analysis, we determine that the acceptable values for $x$ are:

$x > \frac{1 + \sqrt{7}}{2}$ OR $x < \frac{1 - \sqrt{7}}{2}$ However, since $x$ must be greater than 2 for both dimensions of the rectangle and triangle to be valid, we conclude with:

$x > 2$.