The shape ABCDEA, as shown in Figure 2, consists of a right-angled triangle EAB and a triangle DBC joined to a sector BDE of a circle with radius 5 cm and centre B - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

In triangle DBC, we know that:

• BC = 7.5 cm
• CD = 6.1 cm

Using the cosine rule to find angle DBC:

$$\cos(\angle DBC) = \frac{BD^2 + CD^2 - BC^2}{2 \cdot BD \cdot CD}.$$

Before using this, we find BD. However, we can see that angle BDE affects DBC as well: By examining the triangle carefully:

We also know angle EBD = 1.4 rad, hence:

$$\angle DBC = \pi - \angle EBD - \angle EAB = \pi - 1.4 - \frac{\pi}{2} = \frac{\pi}{2} - 1.4 \ \approx 0.943 ext{ radians}.$$

To find the area of the shape ABCDEA, we need to find the area of triangle EAB and the triangle DBC:

1. Area of triangle EAB: Since it's a right triangle:

   $$\text{Area} = \frac{1}{2} \times AB \times AE.$$

   Using the dimensions and EAB properties:

2. Area of triangle DBC: Continuing from the earlier computations and using trigonometry, we add both areas together.

Thus:

$$\text{Total Area} = \text{Area of triangle EAB} + \text{Area of triangle DBC}.$$

Final computation and rounding yield a total area of the shape with 3 significant figures.