Date | November 2013 | Marks available | 3 | Reference code | 13N.2.hl.TZ0.8 |
Level | HL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
The diagram below shows a semi-circle of diameter 20 cm, centre O and two points A and B such that AˆOB=θA^OB=θ, where θ is in radians.
Show that the shaded area can be expressed as 50θ−50sinθ.
Find the value of θ for which the shaded area is equal to half that of the unshaded area, giving your answer correct to four significant figures.
Markscheme
A=12×102×θ−12×102×sinθ M1A1
Note: Award M1 for use of area of segment = area of sector – area of triangle.
=50θ−50sinθ AG
[2 marks]
METHOD 1
unshaded area =π×1022−50(θ−sinθ)
(or equivalent eg 50π−50θ+50sinθ) (M1)
50θ−50sinθ=12(50π−50θ+50sinθ) (A1)
3θ−3sinθ−π=0
⇒θ=1.969 (rad) A1
METHOD 2
50θ−50sinθ=13(π×1022) (M1)(A1)
3θ−3sinθ−π=0
⇒θ=1.969 (rad) A1
[3 marks]
Examiners report
Part (a) was very well done. Most candidates knew how to calculate the area of a segment. A few candidates used r=20.
Part (b) challenged a large proportion of candidates. A common error was to equate the unshaded area and the shaded area. Some candidates expressed their final answer correct to three significant figures rather than to the four significant figures specified.