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 7 - 2014 - Paper 1

To find angle DBC, we first use the fact that angle EBD is given as 1.4 radians, and angle EAB is (\frac{\pi}{2}) radians. Hence, we calculate:

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

Calculating this:

$\text{Angle DBC} = 1.5708 - 1.4 = 0.1708$

Rounding to three decimal places, we get:

$\text{Angle DBC} \approx 0.171 \text{ radians}$

To find the area of the shape ABCDEA, we need to combine the area of triangle EAB and sector BDE, along with triangle DBC.

1. Area of triangle EAB: Since EAB is a right triangle: [\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 5 = 12.5 , \text{cm}^2]

2. Area of triangle DBC: The length BC is 7.5 cm and we already calculated angle DBC as approximately 0.171 radians: [\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \times \sin(\text{Angle DBC})] To find the height, we apply the sine function: [\text{Area} = \frac{1}{2} \times 7.5 \times 5 \times \sin(0.171) \approx \frac{1}{2} \times 7.5 \times 5 \times 0.1709 = 6.55 , \text{cm}^2]

3. Total area of shape ABCDEA: [\text{Total Area} = \text{Area of triangle EAB} + \text{Area of sector BDE} + \text{Area of triangle DBC}] [\text{Total Area} = 12.5 + 17.5 + 6.55 = 36.55 , \text{cm}^2] Rounding to three significant figures gives: 36.6 cm².