Foundation November 2020 Paper 2 Q20
20.
This pattern is made from a regular seven-sided polygon surrounded by squares and isosceles triangles.
Show that the value of \(x\) is 64·3 correct to 1 decimal place. [4]
You must show all your working.
| Scheme | Marks | Comments |
|---|---|---|
| (Interior angle of the heptagon =) 180 – 360 ÷ 7 OR (7 – 2) × 180 ÷ 7 OR (7 × 180 – 360) ÷ 7 | M1 | |
| =128.6(°) or 128.57(…)(°) | A1 | May be seen on diagram. |
| (Unique angle in triangle =) (360 – 90 – 90 – 128.6 =) 51.4(28…°) | B1 | FT ‘their derived 128.6’ May be seen on diagram. |
| Working to show that \(x\) = 64.3 to 1 d.p. (180 – 51.4(28…)) ÷ 2 = 64.285 to 64.3 | B1 | CAO |
| (4) |
Alternative
Alternative method 1 working from 64.3
| Scheme | Marks | Comments |
|---|---|---|
| (Unique angle in triangle =) (180 – 64.3 – 64.3) = 51.4 | B1 | |
| (Interior angle of the heptagon =) (360 - 90 - 90 - 51.4) =128.6 | B1 | FT ‘their 180 – 64.3 – 64.3’ Only awarded if this is clearly the interior angle of the heptagon |
| (Interior angle of the heptagon = ) 180 – (360 ÷ 7) OR (7 – 2) × 180 ÷ 7 OR (7 × 180 – 360) ÷ 7 | M1 | |
| =128.6(°) or 128.57(…)(°) | A1 |
Alternative method 1a for final 2 marks
| Scheme | Marks | Comments |
|---|---|---|
| (Sum of the interior angles of a heptagon=) (7 – 2) × 180 o.e AND 128.6 × 7 | M1 | M0 for ‘their 128.6 × 7’ = 900(.2) alone |
| 900 | A1 | Allow for 900 and 900.2 |
Alternative method 2 using exterior angles
| Scheme | Marks | Comments |
|---|---|---|
| Exterior angle (of the heptagon) = 360 ÷ 7 | M1 | Method must be seen |
| = 51.4(28…°) | A1 | |
| (Unique angle in triangle = ) (360 – 90 – 90 – (180 - 51.4(28…°))) = 51.4(28…°) | B1 | May be seen on diagram. FT ‘their derived 51.4(28…) |
| Working to show that (\(x\) =) (180 – 51.4(28…))÷ 2 = 64.3 | B1 | May be seen on diagram. CAO |