Higher June 2017 Paper 2 Q25
25 The diagram shows a logo.
- ABE and DCE are congruent triangles.
- BCE is a sector of a circle, centre E.
Not drawn accurately
Show that the area of the logo is 510 cm\(^2\) to 2 significant figures. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(15^2 + 26^2 - 2 \times 15 \times 26 \times \cos 38\) or [286, 286.4] or [16.9, 17] | M1 | May be seen in a square root May be seen on diagram |
| \(\dfrac{108}{360}\) or 0.3 or \(\dfrac{360}{108}\) or 3.33(…) | M1 | oe eg \(108 \div 360\) or 30% May be seen in two steps eg \(\times 108 \div 360\) |
| their \(\dfrac{108}{360} \times \pi \times\) [286, 286.4] or \(\pi \times\) their [286, 286.4] \(\div\) their \(\dfrac{360}{108}\) or [269, 272.4114] | M1dep | dep on 1st and 2nd M1 oe eg \(\dfrac{108}{360} \times \pi \times (\text{their } [16.9, 17])^2\) |
| \((2 \times)\ \dfrac{1}{2} \times 15 \times 26 \times \sin 38\) or [120, 120.1] or [240, 240.2] | M1 | oe |
| [509, 512.6114] and 510 | A1 | Must see a value in range [509, 512.6114] and 510 |
Additional guidance
| \(15 \times 26 \times \sin 38\) scores 4th M1 unless subsequently doubled | |
| If (sector) 270 and (2 triangles) 240 followed by \(270 + 240 = 510\) | M4A1 |
| Working back from 510. Apply scheme but maximum mark is M4A0 | |
| Assuming angle AEB = 72 and then using sine rule to work out BE does lead to area = 510 to 2sf but can score a maximum of M0M1M0M1depA0 \(BE = \dfrac{26}{\sin 72} \times \sin 38 = 16.8\) (or 17) \(\dfrac{108}{360} \times \pi \times 16.8^2 = 266 \qquad 2 \times \dfrac{1}{2} \times 15 \times 26 \times \sin 38 = 240.2\) \(506.2 \to 510\) | M0 M1M0depM1 A0 |
| BE = [16.9, 17] seen with no working scores first M1 (and possibly all other marks) | |
| \(BE = 35 \div 2 = 17.5 \to 17\) does not score first M1 |