June 2017 Paper 2 Q18
18 The diagram shows a rectangle with a diagonal drawn.
The given expressions for the measurements are in centimetres.
Not drawn accurately
Work out an expression for the area of the rectangle, in cm\(^2\)
Give your answer in its simplest form, in terms of \(y\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + (2x)^2 = (4y)^2\) or \(x^2 + 4x^2 = 16y^2\) | M1 | oe eg \(5x^2 = (4y)^2\) Missing brackets may be recovered |
| \(x^2 = \dfrac{16}{5}y^2\) or \(5x^2 = 16y^2\) or \(x = \sqrt{\dfrac{16y^2}{5}}\) or \(x\sqrt{5} = 4y\) | M1 | oe equation of the form \(ax^2 = by^2\) or \(cx = dy\) or \(x = \sqrt{ky^2}\) eg \(x^2 = 16y^2 \div 5\) ft if Pythagoras used with only error being missing brackets |
| \(2 \times\) their \(\dfrac{16}{5}y^2\) or their \(16y^2 \div \dfrac{\text{their } 5}{2}\) | M1dep | oe \(2 \times\) their \(x^2\) or \(2 \times\) (their \(x\))\(^2\) dep on at least one M |
| \(\dfrac{32}{5}y^2\) or \(6\dfrac{2}{5}y^2\) or \(6.4y^2\) | A1 |
Additional guidance
| \(5x^2 = (4y)^2\) with no further work | M1M0M0 |
| \(x^2 + 4x^2 = 4y^2\) with answer \(\dfrac{8}{5}y^2\) | M0M1M1A0 |
| \(x^2 + 2x^2 = 16y^2\) with answer \(\dfrac{32}{3}y^2\) | M0M1M1A0 |
| \(x^2 + 2x^2 = 4y^2\) with answer \(\dfrac{8}{3}y^2\) | M0M1M1A0 |
| \((2x)^2 = (4y)^2 - x^2\) \(x^2 = 16y^2\) and \(2 \times 16y^2 = 32y^2\) | M1 M0M1A0 |
| \(\dfrac{32}{5}y^2\) followed by further work | M3A0 |