Foundation November 2019 Paper 3 Q13
13 The midpoints of the sides of a rectangle are joined by straight lines as shown.
Work out the percentage of the rectangle that is shaded. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 75 cao nfww | 4 | M1 for inventing a length and width and correct answer to their length × their width M1 for correct area of one triangle M1 for their rectangle area – 2 × their triangle area oe OR M1 for subdividing shape into right triangles and/or rectangles B2 for shaded area = \(\dfrac{6}{8}\) oe of rectangle or B1 for one triangle = \(\dfrac{1}{8}\) oe or 12.5% of rectangle oe OR M1 for recognising two triangles = rectangle B2 for shaded area = \(\dfrac{3}{4}\) or oe \(\dfrac{6}{8}\) of rectangle or M1 for two triangles = \(\dfrac{1}{4}\) or \(\dfrac{2}{8}\) oe or 25% of rectangle | May be algebraic “x by y” rectangle (Diagram is 11 cm by 5 cm) Accept equal length and width Or a trapezium = half shaded area May be 6 × one triangle or 2 × one trapezium e.g.  May be as 8 triangles make the rectangle May be as 8 triangles or 4 rectangles make the rectangle Example for 11 by 5 M1 for 11 × 5 = 55 M1 for 5.5 × 2.5 ÷ 2 = 6.875 M1 for 55 – 13.75 = 41.25 |