Figure 1 is a sketch representing the cross-section of a large tent ABCDEF - Edexcel - A-Level Maths Pure - Question 5 - 2016 - Paper 2

To calculate the area of the sector FBCD, we can use the formula:

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

Substituting the known values:

• $r = 3.5$ m,
• $\theta = 1.77$ radians,

we get:

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

Thus, the area of the sector FBCD is 10.84 m².

To find the total area of the cross-section of the tent, we combine the area of the sector FBCD with the area of triangle BFD.

1. Area of triangle BFD: We can use the formula:

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

   The base is $BF = 3.5$ m, and we need to find the height which can be calculated using sine:

   • The angle BFD is 1.77 radians. Thus, the height is:

   $h = AF \sin(angle BFD) = 3.7 \times \sin(1.77) \approx 3.7 \times 0.969 = 3.59$

   The area is then:

   $A_{triangle} = \frac{1}{2} \times 3.5 \times 3.59 \approx 6.27 ext{ m}^2$

2. Total Area:

   Combining both areas:

   $Total \: Area = Area_{sector} + Area_{triangle} \approx 10.84 + 6.27 = 17.11 ext{ m}^2$

Rounding to two decimal places, the total area of the cross-section of the tent is approximately 17.11 m².