(a) [Figure 7.1] The diagram shows part of an n-sided regular polygon ABCDEFGH... DCF = EFC = 24°.
- (a)(i)Find the value of n.[2]
- (a)(ii)Find HFG.[2]
- (b)(i)Show that triangle PQS is similar to triangle TQX. Give a reason for each statement you make.[3]
- (b)(ii)Find the ratio SX: SQ.[1]
- (b)(iii)Find the ratio area of triangle TQX: area of parallelogram PQRS.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [2] B1 for exterior angle = 24° or interior angle = 156°; or M1 for 360/24 soi or 180(n-2)/n = 180-24 oe; Final answer n=15.
- [2] M1 for x + x + 156 = 180 oe; Final answer 12°.
- [3] B2 for two correct pairs of angles with correct reasons; or B1 for one pair with a correct reason or two correct pairs of angles identified with wrong/no reasons; ∠Q is common / shared / same (AA similarity); ∠PSQ = ∠TXQ (corresponding angles, TU || QR); ∠SPQ = ∠XTQ (corresponding angles, TU || QR); Hence, triangle PQS is similar to triangle TQX.
- [1] 2:3.
- [2] B1 for [1/2 * (1/3) * (1/3)] : 1 oe (considering ratio of areas of triangles with common height or using ratio of sides squared for similar triangles); After 0 scored SC1 for (1/3)^2 : 1; Final answer 1:18.
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