Higher November 2018 Paper 3 Q16
16 Here is a shaded shape \(ABCD\).
The shape is made from a triangle and a sector of a circle, centre \(O\) and radius 6 cm.
\(OCD\) is a straight line.
\(AD = 14\) cm
Angle \(AOD = 140^\circ\)
Angle \(OAD = 24^\circ\)
Calculate the perimeter of the shape.
Give your answer correct to 3 significant figures. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 39.9 | P1 | for finding the length of the minor or major arc eg \(\dfrac{220}{360} \times \pi \times 12\) (= 23(.03834..)) |
| P1 | for substituting into the sine or cosine rule to find \(OD\) eg \(14 \div \sin 140 = OD \div \sin 24\) or (\(OD^2 =\)) \(6^2 + 14^2 - 2 \times 6 \times 14 \times \cos 24\) (=78.5....) | |
| P1 | for a complete process to find the length \(OD\) eg \(14 \div \sin 140 \times \sin 24\) (=8.8(58778..)) | |
| P1 | for a complete process to find the perimeter eg \(\text{``}23(.03834..)\text{''} + 14 + \text{``}8.8(58778..)\text{''} - 6\) | |
| A1 | for an answer in the range 39.8 to 40 |
Additional guidance
Allow appropriate rounding if calculation seen in parts
Must involve \(OD\) in the relationship but may be implied
May be seen in multiple calculations
If an answer in the range is seen in working and then incorrectly rounded award full marks.