November 2021 Paper 1 Q21
21 \(ABC\) is a triangle.
The perpendicular from \(A\) meets \(BC\) at \(D\).
\(BC = (6 + 2\sqrt{7}\,)\) cm
Not drawn accurately
Area of triangle \(ABC = (13 + 3\sqrt{7}\,)\) cm\(^2\)
Work out the length, in cm, of \(AD\).
Give your answer in the form \(\quad a + b\sqrt{c} \quad\) where \(a\), \(b\) and \(c\) are integers. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Works out \(\dfrac{1}{2} \times (6 + 2\sqrt{7}\,)\) | ||
| \(\dfrac{1}{2} \times (6 + 2\sqrt{7}\,) \times AD\) | M1 | oe eg \((3 + \sqrt{7}\,) \times AD\) or \((3 + \sqrt{7}\,) \times AC \sin C\) may be implied |
| \(\dfrac{13 + 3\sqrt{7}}{3 + \sqrt{7}} \times \dfrac{3 - \sqrt{7}}{3 - \sqrt{7}}\) | M1 | ft their \(\dfrac{13 + 3\sqrt{7}}{3 + \sqrt{7}}\) their denominator must have 2 terms |
| (numerator =) \(39 - 13\sqrt{7} + 9\sqrt{7} - 21\) or \(18 - 4\sqrt{7}\) | M1dep | ft their numerator which must have 2 terms oe dep on 2nd M1 |
| (denominator =) 2 | M1dep | ft their denominator dep on 2nd M1 |
| \(9 - 2\sqrt{7}\) | A1 | |
| Alternative method 2 Works out \(2 \times (13 + 3\sqrt{7}\,)\) | ||
| \(\dfrac{1}{2} \times (6 + 2\sqrt{7}\,) \times AD\) | M1 | oe eg \((6 + 2\sqrt{7}\,) \times AD = (26 + 6\sqrt{7}\,)\) may be implied |
| \(\dfrac{26 + 6\sqrt{7}}{6 + 2\sqrt{7}} \times \dfrac{6 - 2\sqrt{7}}{6 - 2\sqrt{7}}\) | M1 | ft their \(\dfrac{26 + 6\sqrt{7}}{6 + 2\sqrt{7}}\) their denominator must have 2 term |
| (numerator =) \(156 - 52\sqrt{7} + 36\sqrt{7} - 84\) or \(72 - 16\sqrt{7}\) | M1dep | ft their numerator which must have 2 terms oe dep on 2nd M1 |
| (denominator =) 8 | M1dep | ft their denominator dep on 2nd M1 |
| \(9 - 2\sqrt{7}\) | A1 | |
| Alternative method 3 Using identities | ||
| \((6 + 2\sqrt{7}\,) \times AD = (26 + 6\sqrt{7}\,)\) | M1 | oe |
| \((6 + 2\sqrt{7}\,) \times (a + b\sqrt{7}\,) = (26 + 6\sqrt{7}\,)\) | M1 | oe |
| \(6a + 14b = 26\) and \(2a + 6b = 6\) | M1dep | oe eg \(3a + 7b = 13\) and \(a + 3b = 3\) |
| \(a = 9\) or \(b = -2\) | M1dep | |
| \(9 - 2\sqrt{7}\) | A1 | |
Additional guidance
| Alt1 \(\dfrac{18 - 4\sqrt{7}}{2}\) | M4 |
| Alt2 \(\dfrac{72 - 16\sqrt{7}}{8}\) | M4 |
| Omission of \(\dfrac{1}{2}\) can score up to M0M1M1M1A0 eg \(\dfrac{13 + 3\sqrt{7}}{6 + 2\sqrt{7}} \times \dfrac{6 - 2\sqrt{7}}{6 - 2\sqrt{7}}\) \(\dfrac{78 - 26\sqrt{7} + 18\sqrt{7} - 42}{8}\) \(4.5 - \sqrt{7}\) (If the \(\dfrac{1}{2}\) is recovered then all 5 marks are possible) | M0M1 M1M1 A0 |
| Missing brackets must be recovered |