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 center B - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 1

Since angle EBD = 1.4 radians and angle EAB = ( \frac{\pi}{2} ) radians, angle DBC can be calculated as:

$\text{ angle DBC} = \text{ angle EAB} - \text{ angle EBD} = \frac{\pi}{2} - 1.4.$

Calculating this gives:

$\text{ angle DBC} \approx 0.943 \text{ radians.}$

To find the area of shape ABCDEA, we need to add the areas of triangle EAB and triangle CBD to the area of the sector BDE.

1. Calculate the area of triangle EAB:

   • Using the formula for the area of a right-angled triangle, ( \text{Area} = \frac{1}{2} \times AB \times AE ) where ( AB = 5 ) cm and ( AE = 5 \sin(1.4) ):
   • From calculations, ( AE \approx 5 \sin(1.4) \approx 5 \times 0.985 = 4.925 ). Thus:

   $\text{Area EAB} = \frac{1}{2} \times 5 \times 4.925 \approx 12.3125 \text{ cm}^2$.

2. Calculate the area of triangle DBC:

   • The base BC = 7.5 cm and height = 5 \sin(\text{angle DBC}) = 5 \sin(0.943) \approx 5 \times 0.8 = 4.0. Thus:

   $\text{Area DBC} = \frac{1}{2} \times 7.5 \times 4.0 = 15.0 \text{ cm}^2$.

3. Total area of shape ABCDEA:

   $\text{Total Area} = \text{Area EAB} + \text{Area DBC} + \text{Area Sector} \approx 12.3125 + 15.0 + 17.5 \approx 44.8125 \text{ cm}^2 \approx 44.8 \text{ cm}^2$.