Figure 1 is a sketch representing the cross-section of a large tent ABCDEF - Edexcel - A-Level Maths Pure - Question 4 - 2016 - Paper 2

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

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

where:

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

Calculating the area:

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

Thus, the area of sector FBCD is approximately $10.84$ m² (to 2 decimal places).

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

1. Area of triangle AFB: The formula for the area of a triangle is: $A_{triangle} = \frac{1}{2} \times base \times height$ Here, the base is $AF + BF = 3.7 + 3.5 = 7.2$ m and the height is from F perpendicular to AB (which is $\sin(1.77)$): $A_{triangle} = \frac{1}{2} \times 7.2 \times 3.5 \times \sin(1.77)$ Calculating, $A_{triangle} \approx \frac{1}{2} \times 7.2 \times 3.5 \times 0.999$ (since $heta = 1.77$ radians is close to $90°$) $A_{triangle} \approx 12.46$

2. Total Area: Therefore, the total area is: $Total Area = A_{triangle} + A_{sector}$ $Total Area \approx 10.84 + 12.46 \approx 19.30$

Thus, the total area of the cross-section of the tent is approximately $19.30$ m² (to 2 decimal places).