June 2017 Paper 2 Q10
10 \(ABC\) is an isosceles triangle with \(AB = AC\)
The area of \(ABC\) is 57.76 cm\(^2\)
Not drawn accurately
Work out the length of \(AB\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(\dfrac{1}{2} \times x \times x \times \sin 150\) or \(\dfrac{1}{4}x^2\) or \(\dfrac{1}{2} \times b \times c \times \sin 150 = 57.76\) or \(\dfrac{1}{4} \times b \times c = 57.76\) | M1 | oe Any letter(s) |
| \(x^2 = \dfrac{57.76 \times 2}{\sin 150}\) or \(x^2 = 57.76 \times 4\) or \(x^2 = 231(.04)\) or \(\dfrac{1}{2}x = \sqrt{57.76}\) or \(\sqrt{231(.04)}\) or \(2\sqrt{57.76}\) | M1dep | oe eg \(x^2 = \dfrac{57.76}{\frac{1}{2}\sin 150}\) Must have either \(x^2 =\) or \(\dfrac{1}{2}x = \sqrt{57.76}\) or \(\sqrt{231(.04)}\) or \(2\sqrt{57.76}\) Any letter |
| 15.2 | A1 | |
| Alternative method 2 | ||
| \(\dfrac{1}{2} \times x \times x\cos\dfrac{150}{2} \times \sin\dfrac{150}{2}\) \(= \dfrac{57.76}{2}\) | M1 | oe Any letter |
| \(x^2 = \dfrac{57.76}{\cos\frac{150}{2}\sin\frac{150}{2}}\) or \(x^2 = 231(.04)\) or \(\sqrt{231(.04)}\) or \(2\sqrt{57.76}\) | M1dep | oe Must have either \(x^2 =\) or \(\sqrt{231(.04)}\) or \(2\sqrt{57.76}\) Any letter |
| 15.2 | A1 | |
Additional guidance
Do not allow 15 as a misread of 150
\(x\) can be \(b\) or \(AB\) or \(AC\) etc
\(b\) and \(c\) can be \(a\) and \(b\) or \(AB\) and \(AC\) etc