FIGURE I shows a ramp leading to the entrance of a building - NSC Mathematics - Question 8 - 2022 - Paper 2

Using the Cosine Rule in triangle ABE:

$BE^2 = AB^2 + AE^2 - 2(AB)(AE)\cos(\angle BAE)$

Substituting the values we know:

$BE^2 = (1.93)^2 + (0.915)^2 - 2(1.93)(0.915)\cos(120°)$

Calculating each term:

• $(1.93)^2 \approx 3.72$
• $(0.915)^2 \approx 0.84$
• $-2(1.93)(0.915)(-0.5) \approx 0.88$

Thus, putting it together gives:

$BE^2 \approx 3.72 + 0.84 + 0.88 \approx 5.44$

So, taking the square root:

$BE \approx 2.52 \text{ m}$

JFollowing the area formula for a triangle:

$\text{Area} = \frac{1}{2} \times (BF)(FD) \sin(\angle BFD)$

Substituting values into the equation where BF = FD and BF = \frac{5}{7} BE:

Letting (BF = \frac{5}{7} * 2.52 = 1.80 \text{ m}$$

So, the area calculation becomes:

$\text{Area}_{ABFD} = \frac{1}{2} \times (1.80)(1.80) \sin(75°)$

Calculating this:

1. $\sin(75°) \approx 0.966$

Thus:

$\text{Area} \approx \frac{1}{2} \times (1.80)^2 \times 0.966 \approx 1.56 \text{ m}^2$