Photo AI

The diagram shows a sector AOB of a circle with centre O and radius r cm - AQA - A-Level Maths Mechanics - Question 5 - 2017 - Paper 1

Question icon

Question 5

The-diagram-shows-a-sector-AOB-of-a-circle-with-centre-O-and-radius-r-cm-AQA-A-Level Maths Mechanics-Question 5-2017-Paper 1.png

The diagram shows a sector AOB of a circle with centre O and radius r cm. The angle AOB is θ radians. The sector has area 9 cm² and perimeter 15 cm. 5 (a) Show th... show full transcript

Worked Solution & Example Answer:The diagram shows a sector AOB of a circle with centre O and radius r cm - AQA - A-Level Maths Mechanics - Question 5 - 2017 - Paper 1

Step 1

Show that r satisfies the equation 2r² - 15r + 18 = 0.

96%

114 rated

Answer

To show that r satisfies the equation, we need to first use the given formulas for the area and perimeter of a sector.

  1. Area of the Sector: The area A of a sector is given by the formula: A=12r2θA = \frac{1}{2} r^2 \theta Given that the area is 9 cm²: 12r2θ=9\frac{1}{2} r^2 \theta = 9 Rearranging gives: r2θ=18(Equation 1)r^2 \theta = 18\quad\text{(Equation 1)}

  2. Perimeter of the Sector: The perimeter P of a sector is the sum of the lengths of the two radii and the arc length: P=2r+rθP = 2r + r\theta Given that the perimeter is 15 cm: 2r+rθ=15(Equation 2)2r + r\theta = 15\quad\text{(Equation 2)}

  3. Eliminating θ: From Equation 1, we can express θ as: θ=18r2\theta = \frac{18}{r^2} Substituting this expression for θ into Equation 2: 2r+r(18r2)=152r + r \left( \frac{18}{r^2} \right) = 15 Simplifying this leads to: 2r+18r=152r + \frac{18}{r} = 15 Multiplying through by r to eliminate the fraction gives: 2r2+18=15r2r^2 + 18 = 15r Rearranging results in: 2r2−15r+18=0,2r^2 - 15r + 18 = 0, which proves the equation.

Step 2

Find the value of θ. Explain why it is the only possible value.

99%

104 rated

Answer

To find the value of θ, we first need to solve the quadratic equation derived in part (a):

  1. Solving the Quadratic Equation: Using the quadratic formula: r=−b±b2−4ac2ar = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a = 2, b = -15, and c = 18: r=15±(−15)2−4⋅2⋅182⋅2r = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2} Calculate the discriminant: r=15±225−1444r = \frac{15 \pm \sqrt{225 - 144}}{4} r=15±814r = \frac{15 \pm \sqrt{81}}{4} r=15±94r = \frac{15 \pm 9}{4} This results in:

    • r=244=6r = \frac{24}{4} = 6
    • r=64=1.5r = \frac{6}{4} = 1.5

  2. Finding θ: Using the value of r = 6: From Equation 1: r2θ=18  ⟹   62θ=18  ⟹   36θ=18  ⟹   θ=12r^2 \theta = 18\implies\ 6^2 \theta = 18\implies\ 36θ = 18\implies\ θ = \frac{1}{2}

    Using the value of r = 1.5: r2θ=18  ⟹   (1.5)2θ=18  ⟹   2.25θ=18  ⟹   θ=8r^2 \theta = 18\implies\ (1.5)^2 \theta = 18\implies\ 2.25θ = 18\implies\ θ = 8 Since θ must be less than 2π for it to make sense geometrically, only θ = \frac{1}{2} is valid. Hence the only possible value for θ is ( \frac{1}{2} ) radians, because the other value results from a less practical radius for the context.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;