Higher November 2021 Paper 4 Q19
19 ABC is an isosceles triangle.
The sides of the triangle ABC are all tangents to a circle of radius 6 cm, centre O.
Not to scale
Angle BAC = 70° and BA = BC.
(a) Show that length BO is 17.54 cm, correct to 2 decimal places. [4]
(b) Find the area of triangle ABC.
You must show your working. [5]
You must show your working. [5]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| complete correct argument e.g. | accept any correct method not using 17.54 | ||
| angle ABC = 40° | B1 | could be on diagram and also accept ABO = 20°, BO’T’ = 70° | |
| [BO = ] e.g. \(\dfrac{6}{\sin 20}\) | M2 | M1 for e.g \(\sin 20 = \dfrac{6}{[BO]}\) | i.e BO as subject for M2 and condone sine rule with sin 90° for M2 |
| 17.542… or 17.543 | A1dep | dep. on at least M1 | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 202 or 201.5 to 201.8 with correct working | 5 | Accept any correct method e.g. M1 for [height=] 17.54 + 6 or 23.54… M2 for [half base =] \(\dfrac{6}{\tan 35}\) or \(\dfrac{23.54}{\tan 70}\) or M1 for \(\tan 35 = \dfrac{6}{\textit{half base}}\) or \(\tan 70 = \dfrac{23.54}{\textit{half base}}\) | “Correct working” requires evidence of at least M2 or M1M1 Condone use of 8.6 leading to an answer of 202.4… M2 implied by e.g. 8.56 to 8.57 or 17.1 to 17.2 |
| M1 for \(\frac{1}{2}\) × their base × their height oe If 0 scored SC2 for 202 or 201.6 to 201.8 with no working or SC1 for 8.56 to 8.57 or 17.1 to 17.2 with no working | e.g. \(\frac{1}{2}\) × (2 × 8.568…) × 23.54… | ||