4MA0 Higher June 2013 Paper 3 Q16
16
Diagram NOT accurately drawn
Triangle \(ABC\) is similar to triangle \(PQR\).
\(AB\) corresponds to \(PQ\).
\(AC\) corresponds to \(PR\).
\(AB = 8\) cm.
\(AC = 5\) cm.
\(PQ = 14\) cm.
(a) Calculate the length of \(PR\). (2)
The area of triangle \(ABC\) is 16 cm2
(b) Calculate the area of triangle \(PQR\). (3)
| Scheme | Marks |
|---|---|
| \(\dfrac{PR}{5} = \dfrac{14}{8}\) or \(\dfrac{PR}{14} = \dfrac{5}{8}\) | M1 |
| 8.75 | A1 |
| (2) |
Notes
M1: or for \(5 \times \dfrac{14}{8}\) oe
| Scheme | Marks |
|---|---|
\(\dfrac{14}{8}\) or \(\dfrac{7}{4}\) or 1.75 or \(\dfrac{8}{14}\) or \(\dfrac{4}{7}\) or 0.571... (May be implied by second M1) Allow ratio notation | M1 |
| \(1.75^2\) oe eg 3.0625, \(\dfrac{49}{16}\) or \(\left(\dfrac{4}{7}\right)^2\) oe eg \(\dfrac{16}{49}\), 0.326... allow ratio notation | M1 |
| 49 | A1 |
| (3) |
Notes
A1: cao
SC : B1 for an answer of 28
Alternative method
M1 for \(\dfrac{1}{2} \times 8 \times 5 \times \sin A\) and \(\sin A = 0.8\)
M1 (dep) for \(\dfrac{1}{2} \times 14 \times \text{“}8.75\text{”} \times 0.8\)