4MA0 Higher January 2018 Paper 4R Q19
19 The four angles, in degrees, of quadrilateral \(ABCD\) are
angle \(A = (x^2 - 105)\)
angle \(B = (x^2 - 65)\)
angle \(C = (470 - 30x)\)
angle \(D = (510 - 30x)\)
Show that \(ABCD\) is a trapezium.
Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| Eg \(x^2 - 105 + x^2 - 65 + 470 - 30x + 510 - 30x = 360\) or \(2x^2 - 60x + 810 = 360\) | M1 |
| Eg \(2x^2 - 60x + 450\;(= 0)\) or \(2x^2 - 60x = -450\) or \(x^2 - 30x + 225\;(= 0)\) | M1 |
Eg \((x - 15)(x - 15)\;(= 0)\) or \(\dfrac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 225}}{2 \times 1}\) | M1 |
| \(x = 15\) | A1 |
| M1 | |
| 160 and 20 or 120 and 60 with conclusion | A1 |
| (6) | |
| (6 marks) |
Notes
M1: For a correct equation
M1: For a correct three term quadratic
M1: For \((x - 15)(x - 15)\;(= 0)\) or \(2(x - 15)(x - 15)\;(= 0)\) or \((2x - 30)(x - 15)\;(= 0)\) or \(\dfrac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 225}}{2 \times 1}\) oe
(may be partially evaluated; Condone lack of brackets)
A1: Dep on first 2 method marks
M1: For substitution of \(x = 15\) into \(x^2 - 65\) and \(470 - 30x\) or \(x^2 - 105\) and \(510 - 30x\)
Alternative
| Scheme | Marks |
|---|---|
| Eg \(x^2 - 65 + 470 - 30x = 180\) or \(x^2 - 105 + 510 - 30x = 180\) | M1 |
| \(x^2 - 30x + 225\;(= 0)\) or \(x^2 - 30x = -225\) | M1 |
| \((x - 15)(x - 15)\;(= 0)\) or \(\dfrac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 225}}{2 \times 1}\) | M1 |
| \(x = 15\) | A1 |
| M1 | |
| 160 and 20 or 120 and 60 with conclusion | A1 |
Notes
M1: For a correct equation
M1: For a correct three term quadratic
M1: For \((x - 15)(x - 15)\;(= 0)\) or \(\dfrac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times 225}}{2 \times 1}\) oe
(may be partially evaluated; Condone lack of brackets)
A1: Dep on first 2 method marks
M1: For a substitution of \(x = 15\) into the other pairs of co-interior angles.
A1: cso