Higher June 2022 Paper 1R Q17
17 Here is a parallelogram \(PQRS\), in which angle \(SPQ\) is acute.
Diagram NOT accurately drawn
\(PQ = 6.1\) cm \(PS = 3.8\) cm
The area of the parallelogram is 18 cm2
Work out the length of \(QS\)
Give your answer correct to 3 significant figures.
(5)
| Scheme | Marks |
|---|---|
(area \(PQS\) =) \(\dfrac{1}{2} \times 6.1 \times 3.8 \times \sin P = 9\) or (area \(PQRS\) =) \(6.1 \times 3.8 \times \sin P = 18\) or \(\dfrac{1}{2} \times 6.1 \times SX = 9\) or \((SX =)\;\dfrac{9}{\frac{1}{2} \times 6.1}\) (= 2.95…) or \(6.1 \times SX = 18\) or \((SX =)\;18 \div 6.1\) (= 2.95…) | M1 |
eg (\(\sin P\) =) \(\dfrac{9}{\frac{1}{2} \times 6.1 \times 3.8}\left(= 0.776\ldots\text{ or }\dfrac{900}{1159}\right)\) or (\(\sin P\) =) \(\dfrac{18}{6.1 \times 3.8}\left(= 0.776\ldots\text{ or }\dfrac{900}{1159}\right)\) or \((PX^2 =)\;3.8^2 - \text{``}{2.95\ldots}\text{''}^2\) (= 5.73…) or \((PX =)\;\sqrt{3.8^2 - \text{``}{2.95\ldots}\text{''}^2}\) (= 2.39…) | M1 |
(\(P\) =) \(\sin^{-1}\text{``}{0.776\ldots}\text{''}\) (= 50.9…) or \((QX =)\;6.1 - \sqrt{\text{``}{5.73}\text{''}}\) (= 3.70…) or \((QX =)\;6.1 - \text{``}{2.39}\text{''}\) (= 3.70…) | M1 |
\((QS^2 =)\;3.8^2 + 6.1^2 - 2 \times 3.8 \times 6.1 \times \cos(\text{``}{50.9}\text{''})\) (= 22.4…) or \((QS =)\;\sqrt{3.8^2 + 6.1^2 - 2 \times 3.8 \times 6.1 \times \cos(\text{``}{50.9}\text{''})}\) or \((QS^2 =)\;\text{``}{2.95\ldots}\text{''}^2 + \text{``}{3.70\ldots}\text{''}^2\) (= 22.4…) or \((QS =)\;\sqrt{\text{``}{2.95\ldots}\text{''}^2 + \text{``}{3.70\ldots}\text{''}^2}\) | M1 |
| 4.74 | A1 |
| (5) | |
| (5 marks) |
Notes
M1: correct equation for the area of the triangle or parallelogram or a calculation to find the height of the parallelogram (where \(X\) is the point vertically below \(S\) on \(PQ\))
M1: correct expression for \(\sin P\) OR for start of Pythagoras method to find length of \(PX\) (where \(X\) is the point vertically below \(S\) on \(PQ\))
M1: for complete method to find angle \(P\) OR for method to find length of \(QX\)
M1: correct expression for \(QS^2\) (or \(QS\))
A1: accept 4.73 – 4.74