1MA0 Higher March 2013 Paper 2 Q25
25 Here are two triangles \(\mathbf{T}_1\) and \(\mathbf{T}_2\).
Diagram NOT accurately drawn
The lengths of the sides are in centimetres.
The area of triangle \(\mathbf{T}_1\) is equal to the area of triangle \(\mathbf{T}_2\).
Work out the value of \(x\), giving your answer in the form \(a + \sqrt{b}\) where \(a\) and \(b\) are integers. (5)
| Answer | Mark | Notes |
|---|---|---|
| \(1 + \sqrt{5}\) | 5 | M1 for \(\frac{1}{2} \times x \times x \times \sin 30^\circ\) oe M1 for \(\frac{1}{2}(x - 2)(x + 1)\) oe or \(\frac{1}{2} \times (x - 2) \times (x + 1) \times \sin 90\) M1 (dep on at least one previous M1) for formation of equation from equating areas with \(x\) as the only variable A1 for \(x^2 - 2x - 4 = 0\) oe in the form \(ax^2 + bx + c = 0\) or \(ax^2 + bx = c\) A1 cao |