Figure 1 is a sketch representing the cross-section of a large tent ABCDEF - Edexcel - A-Level Maths Pure - Question 4 - 2017 - Paper 3

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

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

where:

• $A$ is the area,
• $r = 3.5 \text{ m}$,
• $\theta = 1.77 \text{ radians}$.

Substituting the values:

$A = \frac{1}{2} \times (3.5)^2 \times 1.77 = \frac{1}{2} \times 12.25 \times 1.77 \approx 10.84 \text{ m}^2$

Thus, the area of the sector FBCD is approximately 10.84 m² when rounded to two decimal places.

To find the total area of the cross-section of the tent, we need to sum the area of the sector FBCD and the area of the triangle AFB.

The area of triangle AFB can be calculated using:

$A_{triangle} = \frac{1}{2} \times base \times height$

Here, base $AF = 3.7 \text{ m}$ and height is the same as the radius $3.5 \text{ m}$. Thus:

$A_{triangle} = \frac{1}{2} \times 3.7 \times 3.5 = \frac{1}{2} \times 12.95 \approx 6.475 \text{ m}^2$

Now combining the area of the triangle and sector:

$Total_Area = 10.84 + 6.475\approx 17.315 \text{ m}^2$

So, the total area of the cross-section of the tent is approximately 17.31 m² rounded to two decimal places.