Foundation November 2019 Paper 3 Q30
30 A is in the shape of a quarter circle of radius 15 cm.
B is in the shape of a circle.
The area of A is 9 times the area of B.
Show that the radius of B is 2.5 cm. (3)
| Answer | Mark | Mark scheme |
|---|---|---|
| Result shown | M1 | for finding the area of A or the area of B, eg \((\pi \times 15^2) \div 4\ (=56.25\pi = 176.(7...)\) or \(177)\) or \(\pi \times 2.5^2\ (= 6.25\pi = 19.6(3...))\) |
| M1 | for finding the area of A and the area of B, eg \((\pi \times 15^2) \div 4\) or \(\text{``}6.25\pi\text{''} \times 9\ (=56.25\pi = 176.(7...)\) or \(177)\) AND \(\pi \times 2.5^2\) or \(\text{``}56.25\pi\text{''} \div 9\ (= 6.25\pi = 19.6(3...))\) | |
| C1 | for conclusion eg, \(\sqrt{56.25\pi \div 9 \div \pi} = 2.5\) oe or \(\sqrt{\dfrac{6.25\pi \times 9 \times 4}{\pi}} = 15\) oe or \(56.25\pi \div 9 = 19.6(3...)\) and \(\pi \times 2.5^2 = 19.6(3...)\) oe or \(6.25\pi \times 9 = 176.(7...)\) or 177 and \((\pi \times 15^2) \div 4 = 176(.7..)\) or 177 oe or for \(((\pi \times 15^2) \div 4) \div (\pi \times 2.5^2) = 9\) oe |
Additional guidance
May work without \(\pi\) or with an approximation of \(\pi\)
Values may be rounded or truncated