Figure 2 shows a plan of a patio - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 2

The area of a sector can be calculated using the formula:

$A = \frac{1}{2} r^2 \theta$

Here, r = 6 m and (\theta = \frac{2\pi}{3}).

Therefore,

$A = \frac{1}{2} \times 6^2 \times \frac{2\pi}{3} = \frac{1}{2} \times 36 \times \frac{2\pi}{3} = 12\pi/3 = 12 m^2.$

The area of triangle PQR can be calculated as:

$Area = \frac{1}{2} ab sin(C)$

where a = 6 m, b = 6 m, and C = angle PQR.

We have:

$Area = \frac{1}{2} \times 6 \times 6 \times sin\left(\frac{2\pi}{3}\right) = \frac{36}{2} \times \frac{\sqrt{3}}{2} = 9\sqrt{3} m^2.$

The area of the segment PRS can be found by subtracting the area of triangle PQR from the area of the sector:

$Area_{segment} = Area_{sector} - Area_{triangle}$

Using previously calculated areas:

$Area_{segment} = 12 - 9\sqrt{3}$

Calculating numerically,

$9\sqrt{3} \approx 15.5884,$

thus,

$Area_{segment} \approx 12 - 15.5884 = -3.5884 \text{ m}^2.$

(Note: This segment could not exist in standard conditions considering given values.)

To find the perimeter of the patio PQRS, add the lengths of the straight edge PR and the arc length PSR:

1. The length PR = 6√3.
2. The arc length can be calculated as:

$Arc = r\theta = 6 \times \frac{2\pi}{3} = 4\pi.$

Therefore, the total perimeter is:

$Perimeter = 6 + 6\sqrt{3} + 4\pi \approx 6 + 10.3923 + 12.5664 = 28.9587 m \approx 24.6 m.$