4MA0 Higher June 2015 Paper 4R Q13
13 \(PQRS\) and \(PLMN\) are similar quadrilaterals.
\(PN = 12\) cm, \(NS = 8\) cm, \(PL = 9\) cm and \(RS = 13.5\) cm.
\(LM\) is parallel to \(QR\) and \(MN\) is parallel to \(RS\).
Diagram NOT accurately drawn
(a) Work out the length of \(MN\). (2)
(b) Work out the length of \(LQ\). (2)
The area of \(PLMN\) is \(A\) cm2
The area of \(PQRS\) is \(kA\) cm2
(c) Find the value of \(k\). (1)
The area of the shaded region is 105.6 cm2
(d) Work out the value of \(A\). (3)
| Scheme | Marks |
|---|---|
Eg. \(\dfrac{12}{12 + 8}\) or \(\dfrac{3}{5}\) or 0.6 or \(\dfrac{12 + 8}{12}\) or \(\dfrac{5}{3}\) or 1.66(66...) or \(\dfrac{MN}{13.5} = \dfrac{12}{12 + 8}\) oe or \((MN =)\;\dfrac{12}{12 + 8} \times 13.5\) oe | M1 |
| 8.1 | A1 |
| (2) |
Notes
M1: for correct scale factor or correct equation involving \(MN\) or correct expression for \(MN\)
Allow use of 1.66(66....) in place of \(\dfrac{5}{3}\) if rounded or truncated to at least 3 significant figures
A1: oe
| Scheme | Marks |
|---|---|
Eg \(\dfrac{PQ}{9} = \dfrac{12 + 8}{12}\) oe or \(\dfrac{PQ}{9} = \dfrac{13.5}{\text{"}8.1\text{"}}\) or \(PQ = 9 \times \dfrac{12 + 8}{12}\) oe or \(PQ = 9 \times \dfrac{13.5}{\text{"}8.1\text{"}}\) oe or \(PQ = 15\) or \(\dfrac{LQ}{9} = \dfrac{8}{12}\) oe or \((LQ =)\;9 \times \dfrac{8}{12}\) oe | M1 |
| 6 | A1 |
| (2) |
Notes
M1: Correct expression for \(PQ\) or \(LQ\) (eg \(9 \times \dfrac{5}{3}\) oe or \(9 \times \dfrac{2}{3}\) oe)
Correct equation involving \(PQ\) or \(LQ\) (eg \(\dfrac{PQ}{9} = \dfrac{5}{3}\) oe or \(\dfrac{LQ}{9} = \dfrac{2}{3}\) oe)
Allow use of 1.66(66....) in place of \(\dfrac{5}{3}\) or 0.666(66...) in place of \(\dfrac{2}{3}\) if rounded or truncated to at least 3 significant figures
| Scheme | Marks |
|---|---|
| \(\dfrac{25}{9}\) oe | B1 |
| (1) |
Notes
B1: Accept \(2\dfrac{7}{9}\)
Accept 2.77(777...) rounded or truncated to at least 3 significant figures
Also accept \(\left(\dfrac{20}{12}\right)^2\) or \(\left(\dfrac{5}{3}\right)^2\)
| Scheme | Marks |
|---|---|
Eg \(\text{"}\left(\dfrac{25}{9}\right)\text{"}A - A = 105.6\) or \(A\left(\text{"}\left(\dfrac{25}{9}\right)\text{"} - 1\right) = 105.6\) or \(A\left(\text{"}\left(\dfrac{16}{9}\right)\text{"}\right) = 105.6\) or \(\text{"}\left(\dfrac{25}{9}\right)\text{"}A = 105.6 + A\) or \(\dfrac{105.6 + A}{\text{"}25/9\text{"}} = A\) | M1 |
| \((A =)\;\dfrac{105.6}{\frac{25}{9} - 1}\) or \((A =)\;\dfrac{105.6}{1.77\ldots}\) or \((A =)\;\dfrac{105.6 \times 9}{16}\) | M1 |
| 59.4 | A1 |
| (3) |
Notes
M1: For a correct equation involving \(A\) ft from part (c)
M1: For correct expression for \(A\).
ft from part (c)
Decimal values should be rounded or truncated correct to at least 3SF
A1: oe