June 2025 Paper 2 Q2
2 \(C\) and \(E\) are points on the circle with equation \(\quad x^2 + y^2 = 144\)
\(CD\) is a straight line.
\(OD : DE = 5 : 1\)
Work out the shaded area.
Give your answer as a decimal. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((r =)\ \sqrt{144}\) or \((r =)\ 12\) | M1 | may be on diagram |
| \(\dfrac{1}{4} \times \pi \times \sqrt{144} \times \sqrt{144}\) or \(36\pi\) or [113, 113.112] or \(\dfrac{1}{2} \times \dfrac{5}{5 + 1} \times \sqrt{144} \times \sqrt{144}\) or 60 | M1dep | oe area of \(\dfrac{1}{4}\) circle eg \(\dfrac{1}{4} \times \pi \times 12 \times 12\) or area of \(COD\) eg \(\dfrac{1}{2} \times 10 \times 12\) |
| \(\dfrac{1}{4} \times \pi \times \sqrt{144} \times \sqrt{144}\) \(- \dfrac{1}{2} \times \dfrac{5}{5 + 1} \times \sqrt{144} \times \sqrt{144}\) or \(36\pi - 60\) | M1dep | oe full method |
| [53, 53.112] | A1 | SC2 [7637.7, 7648.13] or \(5184\pi - 8640\) |
Additional guidance
| Accept [3.14, 3.142] for \(\pi\) throughout | |
| SC2 is from using a radius of 144 | |
| (corrected from the printed mark scheme: the SC2 interval is printed as [7637.7, 7648,13]) |