Higher June 2022 Paper 1 Q19
19
Diagram NOT accurately drawn
\(OAB\) is a sector S of a circle with centre \(O\) and radius \((r + 7)\) metres.
Angle \(AOB = 45^\circ\)
A circle C has radius \((r - 2)\) metres.
The area of sector S is twice the area of circle C
Find the value of \(r\)
Show your working clearly.
(5)
| Scheme | Marks |
|---|---|
| \(\pi \times (r + 7)^2 \times \dfrac{45}{360}\) oe or \((2 \times)\;\pi \times (r - 2)^2\) oe | M1 |
| \(\pi \times (r + 7)^2 \times \dfrac{45}{360} = 2 \times \pi \times (r - 2)^2\) oe | M1 |
E.g. \(675r^2 - 3510r + 675\;(= 0)\) \(15r^2 - 78r + 15\;(= 0)\) oe or \(5r^2 - 26r + 5\;(= 0)\) oe Allow \(5r^2 - 26r = -5\) or \(\left[4(r - 2)\right]^2 = (r + 7)^2\) or \((r - 2)^2 = \left[\dfrac{(r + 7)}{4}\right]^2\) | A1 |
\((5r - 1)(r - 5)\;(= 0)\) oe or \((r =)\;\dfrac{--26 \pm \sqrt{(-26)^2 - 4 \times 5 \times 5}}{2 \times 5}\) or \(5\left(\left(r - \dfrac{26}{10}\right)^2 - \left(\dfrac{26}{10}\right)^2\right) + 5 = 0\) oe or \(4r - 8 = r + 7\) oe | M1 |
| Working required Answer: 5 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for a correct equation
A1: (dep on M2) writing a correct quadratic expression in form \(ax^2 + bx + c\;(= 0)\)
allow \(ax^2 + bx = c\)
M1: (dep on M1) for a complete method to solve their 3-term quadratic equation
Allow one sign error and some simplification – allow as far as
\(\dfrac{26 + \sqrt{676 - 100}}{10}\)
A1: dep on M2
(5 and \(\dfrac{1}{5}\) scores M1M1A1M1A0)