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 the angle DBC, we can use the geometry of the triangle. Since the points A, B, and C are collinear and BC = 7.5 cm:

Let ( x ) be the angle DBC. By using the cosine rule in triangle BDC:

$\cos(x) = \frac{BD^2 + DC^2 - BC^2}{2 \times BD \times DC}$

where:

• ( BD = 5 ) cm (the radius)
• ( DC = 6.1 ) cm
• ( BC = 7.5 ) cm.

Substituting in gives:

$\cos(x) = \frac{5^2 + 6.1^2 - 7.5^2}{2 \times 5 \times 6.1}$

Calculating the individual components:

• ( 5^2 = 25 )
• ( 6.1^2 = 37.21 )
• ( 7.5^2 = 56.25 )

Thus:

$\cos(x) = \frac{25 + 37.21 - 56.25}{2 \times 5 \times 6.1} = \frac{5.96}{61}\approx 0.0977$

Therefore:

$x \approx \cos^{-1}(0.0977) \approx 1.469 \text{ radians}$

To find the area of shape ABCDEA, we can add the areas of triangle EAB, triangle DBC, and sector BDE.

1. Area of triangle EAB: Since it is a right-angled triangle: $\text{Area}_{EAB} = \frac{1}{2} \times AB \times AE$ where (AB = 5) cm and (AE = 5\times \tan(1.4)\approx 5 \times 5.636\approx 28.18 \text{ cm}$Therefore:$\text{Area}_{EAB} = \frac{1}{2} \times 5 \times 28.18 = 70.45 \text{ cm}^2$$

2. Area of triangle DBC: Using the formula for the area of a triangle: $\text{Area}_{DBC} = \frac{1}{2} \times BC \times BD \times \sin(x)$ Here: $\text{Area}_{DBC} = \frac{1}{2} \times 7.5 \times 5 \times \sin(1.469)\approx 18.79 \text{ cm}^2$

3. Total area ABCDEA: $\text{Total Area} = \text{Area}_{EAB} + \text{Area}_{DBC} + \text{Area}_{BDE}\approx 70.45 + 18.79 + 17.5 = 106.74 \text{ cm}^2$

Finally, rounding to 3 significant figures gives: $\text{Total Area} \approx 106 \text{ cm}^2$