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The diagram shows a sector AOB of a circle with centre O and radius r cm - AQA - A-Level Maths: Pure - Question 5 - 2017 - Paper 1
Question 5
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: Pure - Question 5 - 2017 - Paper 1
Step 1
Show that r satisfies the equation 2r² - 15r + 18 = 0.
Answer
To solve this, we first use the formulas for the area and perimeter of a sector.
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Area of the Sector: The area A of a sector is given by the formula:
Given that the area is 9 cm²:
\Rightarrow \theta = \frac{18}{r^2}$$ -
Perimeter of the Sector: The perimeter P of a sector is given by:
Given that the perimeter is 15 cm:
Substituting the expression for \theta:
\Rightarrow 15 = 2r + \frac{18}{r}\ \Rightarrow 15r = 2r^2 + 18\ \Rightarrow 2r^2 - 15r + 18 = 0$$ Hence, we have shown that r satisfies the equation.
Step 2
Find the value of θ. Explain why it is the only possible value.
Answer
To find θ, we solve the quadratic equation derived above:
We apply the quadratic formula:
where a = 2, b = -15, and c = 18:
\Rightarrow r = \frac{15 \pm \sqrt{225 - 144}}{4}\ \Rightarrow r = \frac{15 \pm \sqrt{81}}{4}\ \Rightarrow r = \frac{15 \pm 9}{4}\ \Rightarrow r = \frac{24}{4} = 6\ \ \text{or} \ r = \frac{6}{4} = 1.5\ Now, substituting r = 6 back into the equation for \theta: $$\theta = \frac{18}{6^2} = \frac{18}{36} = \frac{1}{2} \, \text{radians}$$ For r = 1.5: $$\theta = \frac{18}{(1.5)^2} = \frac{18}{2.25} = 8 \, \text{radians}$$ Given that the angle of a sector must be positive and not exceed 2π, the solution for r = 1.5 gives an unusually high theta value which indicates an impracticality for the sector's shape. Thus, the only valid value for θ is \frac{1}{2} radians.