Higher June 2018 Paper 2 Q19
19 A pentagon is made from a square and an isosceles triangle.
Not drawn accurately
Work out the perimeter of the pentagon. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 Using one half of the isosceles triangle | ||
| (base angle =) 35 or (top angle =) 55 | B1 | may be on diagram |
| \(\cos(\text{their } 35) = \dfrac{6}{x}\) or \(\sin(\text{their } 55) = \dfrac{6}{x}\) or \(6^2 + (6 \tan(\text{their } 35))^2\) | M1 | oe eg \(\dfrac{\sin 90}{x} = \dfrac{\sin(\text{their } 55)}{6}\) any letter their 35 must be acute their 55 must be acute |
| \(\dfrac{6}{\cos(\text{their } 35)}\) or \(\dfrac{6}{\sin(\text{their } 55)}\) or \(\sqrt{6^2 + (6 \tan(\text{their } 35))^2}\) or 7.3(2…) | M1dep | oe |
| [50.6, 50.65] | A1ft | ft B0M2 with evaluation of \(36 + 2 \times\) their 7.3(2…) |
| Alternative method 2 Using the isosceles triangle | ||
| (base angle =) 35 or (top angle =) 110 | B1 | may be on diagram |
| \(\dfrac{x}{\sin(\text{their } 35)} = \dfrac{12}{\sin(\text{their } 110)}\) or \(12^2 = x^2 + x^2 - 2 \times x \times x \times \cos(\text{their } 110)\) or \(x^2 = x^2 + 12^2 - 2 \times x \times 12 \times \cos(\text{their } 35)\) | M1 | oe any letter their 35 must be acute their 110 cannot be 125 |
| \(\dfrac{12}{\sin(\text{their } 110)} \times \sin(\text{their } 35)\) or \(\sqrt{\dfrac{12^2}{2 - 2\cos(\text{their } 110)}}\) or \(\dfrac{12^2}{2 \times 12 \times \cos(\text{their } 35)}\) or 7.3(2…) | M1dep | oe |
| [50.6, 50.65] | A1ft | ft B0M2 with evaluation of \(36 + 2 \times\) their 7.3(2…) |
Additional guidance
| Allow B1 even if the angle is not subsequently used | |
| Alt 2 Top angle 90 | M0M0A0 |
| Answer [50.6, 50.65] (possibly from scale drawing) | B1M1M1A1 |