Figure 2 shows the cross-section ABCD of a small shed. The straight line AB is vertical and has length 2.12 m. The straight line AD is horizontal and has length 1.86 m. The curve BC is an arc of a circle with centre A, and CD is a straight line. Given that the size of ∠BAC is 0.65 radians, find (a) the length of the arc BC, in m, to 2 decimal places, (b) the area of the sector BAC, in m², to 2 decimal places, (c) the size of ∠CAD, in radians, to 2 decimal places, (d) the area of the cross-section ABCD of the shed, in m², to 2 decimal places.

A-Level Edexcel Maths: Pure Applications of Differentiation: Figure 2 shows the cross-section

Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths: Pure - Question 10 - 2006 - Paper 2

Question 10

Figure 2 shows the cross-section ABCD of a small shed. The straight line AB is vertical and has length 2.12 m. The straight line AD is horizontal and has length 1.86... show full transcript

Worked Solution & Example Answer:Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths: Pure - Question 10 - 2006 - Paper 2

Step 1

the length of the arc BC, in m, to 2 decimal places

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Answer

To find the length of arc BC, we use the formula:

$l = r \theta$ where:

$r = 2.12 \ m$ (radius)
$\theta = 0.65 \ rad$

Substituting the values:

$l = 2.12 \times 0.65 = 1.378 \ m$

Thus, rounding to 2 decimal places, the length of arc BC is approximately 1.38 m.

Step 2

the area of the sector BAC, in m², to 2 decimal places

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Answer

The area of sector BAC can be calculated using the formula:

$A = \frac{1}{2} r^2 \theta$ Where:

$r = 2.12 \ m$
$\theta = 0.65 \ rad$

Substituting the values:

$A = \frac{1}{2} (2.12)^2 \times 0.65 = 1.4586 \ m²$

Thus, rounding to 2 decimal places, the area of sector BAC is approximately 1.46 m².

Step 3

the size of ∠CAD, in radians, to 2 decimal places

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Answer

We know that

$\text{Since } \theta = \frac{\pi}{2} - \alpha$

Thus, $\alpha = \frac{\pi}{2} - 0.65 \approx 0.92 \ radians$

Therefore, the size of ∠CAD is approximately 0.92 radians.

Step 4

the area of the cross-section ABCD of the shed, in m², to 2 decimal places

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Answer

To find the total area of the cross-section ABCD, we sum the areas of sector BAC and triangle ACD.

Area of Sector BAC: We found this to be 1.46 m².
Area of Triangle ACD: The area can be calculated using:
$= \frac{1}{2} (2.12)(1.86) \sin(\alpha)$$$

Substituting the value of\ \alpha:\n $\alpha ≈ 0.92$ :

$A_{ACD} ≈ \frac{1}{2} (2.12)(1.86) \times 0.81 ≈ 1.57 \ m²$

Summing both areas:

$\text{Total Area} = 1.46 + 1.57 = 3.03 \ m²$

Thus, the area of the cross-section ABCD is approximately 3.03 m².

Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths: Pure - Question 10 - 2006 - Paper 2

Worked Solution & Example Answer:Figure 2 shows the cross-section ABCD of a small shed - Edexcel - A-Level Maths: Pure - Question 10 - 2006 - Paper 2

the length of the arc BC, in m, to 2 decimal places

the area of the sector BAC, in m², to 2 decimal places

the size of ∠CAD, in radians, to 2 decimal places

the area of the cross-section ABCD of the shed, in m², to 2 decimal places

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