Figure 1 is a sketch representing the cross-section of a large tent ABCDEF - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 3

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

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

For this sector:

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

Calculating the area:

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

Calculating further:

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

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

The total area of the cross-section of the tent can be calculated by adding the area of the sector FBCD to the area of triangle AFB. The area of triangle AFB can be calculated using the formula:

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

For triangle AFB:

• The base is AF = 3.7 m.
• The height from F to line AB is calculated using the sine of angle BFD:

$$height = BF \times \sin(angle BFD) = 3.5 \times \sin(1.77)$$

Calculating:

$$sin(1.77) \approx 0.978.$$

Thus,

$$height \approx 3.5 \times 0.978 \approx 3.43 m.$$

Now substituting back:

$$A_{triangle} = \frac{1}{2} \times 3.7 \times 3.43 = 6.35 \text{ m}^2.$$

Finally, adding the area of the sector to the area of the triangle gives:

$$Total Area = 10.84 + 6.35 = 17.19 ext{ m}^2.$$

Thus, the total area of the cross-section of the tent is 17.19 m².