4MA0 Higher June 2013 Paper 3R Q14
14 Triangles \(ABC\) and \(ACD\) are similar.
Diagram NOT accurately drawn
Angle \(BAC\) = angle \(CAD\).
Angle \(ABC\) = angle \(ACD\).
\(AB = 5\) cm and \(AC = 8\) cm.
(a) Calculate the length of \(AD\). (2)
The area of triangle \(ABC\) is 12 cm2
(b) Calculate the area of triangle \(ACD\). (2)
| Scheme | Marks |
|---|---|
| \(8 \times \dfrac{8}{5}\) oe | M1 |
| 12.8 oe | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(12 \times 1.6^2\) | M1 |
| 30.72 | A1 |
| (2) |
Notes
M1: M1 for \(1.6^2\) (=2.56) or \(0.625^2\) (=0.39..)
or \(\left(\dfrac{8}{5}\right)^2 \left(= \dfrac{64}{25}\right)\) or \(\left(\dfrac{5}{8}\right)^2 \left(= \dfrac{25}{64}\right)\)
or 0.5 x 8 x “12.8” x sin 36.9
A1: cao