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The diagram shows two shaded shapes, A and B - Edexcel - GCSE Maths - Question 23 - 2020 - Paper 1

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The diagram shows two shaded shapes, A and B. Shape A is formed by removing a sector of a circle with radius $(3x - 1)$ cm from a sector of the circle with radius $... show full transcript

Worked Solution & Example Answer:The diagram shows two shaded shapes, A and B - Edexcel - GCSE Maths - Question 23 - 2020 - Paper 1

Step 1

Derive an algebraic expression for the area of A

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Answer

The area of the sector A can be calculated using the formula for the area of a sector:

A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2

Here, let the angle be (\theta) (in degrees) for the circle with radius (3x - 1) and the angle for the circle with radius (5 - 1 = 4) cm. The formula results in:

A=θ360×π(3x1)2θ360×π(42)A = \frac{\theta}{360} \times \pi (3x - 1)^2 - \frac{\theta'}{360} \times \pi (4^2)

Step 2

Find the area of shape B

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Answer

Shape B is a circle with diameter ((2 - 2) = 0) cm. The radius is therefore:

rB=02=0r_B = \frac{0}{2} = 0

Thus, the area of shape B is:

AreaB=πrB2=π(0)2=0\text{Area}_B = \pi r_B^2 = \pi (0)^2 = 0

Step 3

Equate the area of shape A to the area of shape B

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Answer

Since the area of shape A is equal to the area of shape B:

A=0A = 0

We can simplify the expression we derived for the area of shape A and set it equal to zero.

Step 4

Solve for x

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Answer

After substituting the expressions and simplifying:

  • Rearranging into a quadratic:

    3x26x+0=03x^2 - 6x + 0 = 0

Then factor or apply the quadratic formula, using the context of the initial formulation to find:[ x = \frac{6 \pm \sqrt{(6)^2 - 4(3)(0)}}{2(3)} = 2 \text{ or } 0 \text{ (invalid, since } x > 1)$$

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