Higher June 2025 Paper 2 Q17
17. A kite is shown.
A logo is created by arranging 8 of these kites inside an octagon.
Diagram not drawn to scale
Calculate the area of one shaded triangle.
Hence, calculate the total area of the shaded region of the logo. [6]
| Scheme | Marks | Comments | ||||||
|---|---|---|---|---|---|---|---|---|
| Note: Do not penalise incorrect units. | ||||||||
| Obtuse angle in a shaded triangle = 152 (°) From a valid method e.g. \((360 - 2\times104)\) OR \((180 - 2\times14)\) = | B3 | If not B3 then B2 for either of the following
B1 for either of the following
| ||||||
| (Area of a shaded triangle =) \(0.5\times11\times11\times\sin(152)\) | M1 | May be seen in stages FT ‘their derived 152’ provided obtuse and \(\ne\) 107 or 135 | ||||||
| = 28.4…(mm2) | A1 | FT May be implied by the final answer | ||||||
| (Area of a shaded region = \(8\times28.4\)…) = 227.2(2…) (mm2) | B1 | FT provided M1 is awarded Allow 227 from correct working.
|
Alternative method using the acute angle of a shaded triangle
| Scheme | Marks | Comments | ||||||
|---|---|---|---|---|---|---|---|---|
| (Acute angle of a shaded triangle =) 14 (°) From a valid method e.g. \((180 - 107 - 45) \div 2\) oe | B2 | B1 for either of the following
| ||||||
| (Area of half a shaded triangle =) \(\dfrac{11\sin(14)\times11\cos(14)}{2}\) oe (\(\frac{1}{2}\times2.66..\times10.67…\)) | M2 | FT ‘their derived 14’ provided acute May be seen in stages | ||||||
| = 14.2…(mm2) | A1 | FT May be seen as area of shaded triangle = 28.4…(mm2) | ||||||
| (Area of a shaded region = \(16\times14.2\)… or \(8\times28.4\)…) = 227.2(2…) (mm2) | B1 | FT ‘their 14.2’ or ‘their 28.4’ provided M2 is awarded Allow 227 from correct working.
|
| (6) |
Corrected by Westie's Workshop
- The board's mark scheme printed “(360 – 107 – 45) ÷ 2)”. Changed to “(360 – 107 – 45) ÷ 2” because an extra closing bracket was printed.
- The board's mark scheme printed “(180 – 107 – 45) ÷ 2)”. Changed to “(180 – 107 – 45) ÷ 2” because an extra closing bracket was printed.
- The board's mark scheme printed “(180 – 107 – 45) ÷ 2)”. Changed to “(180 – 107 – 45) ÷ 2” because an extra closing bracket was printed.
Original: see the board's mark scheme PDF (link above).