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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

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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. ... show full transcript

Worked Solution & Example Answer: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

Step 1

Find, in cm², the area of the sector BDE.

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Answer

To find the area of the sector BDE, use the formula:

Area=12r2θ\text{Area} = \frac{1}{2} r^2 \theta

Where (r = 5) cm and (\theta = 1.4) radians.

Substituting the values, we have:

Area=12×52×1.4=12×25×1.4=17.5 cm2\text{Area} = \frac{1}{2} \times 5^2 \times 1.4 = \frac{1}{2} \times 25 \times 1.4 = 17.5 \text{ cm}^2

Thus, the area of the sector BDE is 17.5 cm².

Step 2

Find the size of the angle DBC, giving your answer in radians to 3 decimal places.

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Answer

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:

Angle DBC=Angle EABAngle EBD=π21.4\text{Angle DBC} = \text{Angle EAB} - \text{Angle EBD} = \frac{\pi}{2} - 1.4

Calculating this:

Angle DBC=1.57081.4=0.1708\text{Angle DBC} = 1.5708 - 1.4 = 0.1708

Rounding to three decimal places, we get:

Angle DBC0.171 radians\text{Angle DBC} \approx 0.171 \text{ radians}

Step 3

Find, in cm², the area of the shape ABCDEA, giving your answer to 3 significant figures.

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Answer

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.

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