In the diagram, the square and the trapezium share a common side of length $x$ cm - OCR - GCSE Maths - Question 12 - 2018 - Paper 6

The area of the trapezium can be calculated using the formula:

$A_{trapezium} = \frac{1}{2} \times (a+b) \times h$

In this case, the two bases of the trapezium are:

• The top base, which is equal to $x$ cm.
• The bottom base, which is fixed at 10 cm.
• The height of the trapezium is 6 cm.

Thus:

$A_{trapezium} = \frac{1}{2} \times (x + 10) \times 6$

Since the area of the square is equal to the area of the trapezium, we can set the two formulas equal to each other:

$x^2 = \frac{1}{2} \times (x + 10) \times 6$

This simplifies to:

$x^2 = 3(x + 10)$

Expanding gives:

$x^2 = 3x + 30$

Rearranging this results in:

$x^2 - 3x - 30 = 0$

To solve the quadratic equation, we can apply the quadratic formula:

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

In this case, $a = 1$, $b = -3$, and $c = -30$. Plugging in these values gives:

$x = \frac{{3 \pm \sqrt{{(-3)^2 - 4 \cdot 1 \cdot (-30)}}}}{2 \cdot 1}$

This simplifies to:

$x = \frac{{3 \pm \sqrt{{9 + 120}}}}{2} = \frac{{3 \pm \sqrt{129}}}{2}$

Calculating the square root, we find:

$x = \frac{{3 \pm 11.36}}{2}$

Calculating these two possible solutions gives:

1. $x = \frac{{3 + 11.36}}{2} = \frac{14.36}{2} \approx 7.18$
2. $x = \frac{{3 - 11.36}}{2} = \frac{{-8.36}}{2} \approx -4.18$

Since $x$ must be a positive length, we take:

$x \approx 7.18$