Chapter 8 Practice Paper: Angles, Triangles and Polygons — mark scheme (print)

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QuizQuipPractice Series

QQ/S1MATH/CH8

Secondary 1 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Angles, Triangles and Polygons

Chapter 8 Practice Paper · Maximum mark: 30

Notes for markers

Award one mark for each point marked [1], up to the marks shown for the part.
Accept answers worded differently that show the same understanding.
Units are required where the answer line shows one. ecf: allow an error carried forward.

Section A: Multiple-choice questions[10 marks]

Question	Answer
1	DComplementary angles add to 90°: 90 - 28 = 62°.	1
2	AAngles on a straight line sum to 180°: 180 - 94 = 86°.	1
3	CVertically opposite angles are equal, so the opposite angle is also 125°. The 55° angles are adjacent, not opposite.	1
4	BCo-interior angles between parallel lines add to 180°: 180 - 72 = 108°.	1
5	D2x + 3x + 4x = 180°, so 9x = 180° and x = 20°.	1
6	AThe exterior angle equals the sum of the two opposite interior angles: 45 + 65 = 110°.	1
7	CFor n = 6: (6 - 2) × 180° = 720°.	1
8	BThe exterior angles sum to 360°, and there are 8 of them: 360 ÷ 8 = 45°.	1
9	AThe exterior angles are all equal and sum to 360°: 360 ÷ 12 = 30 sides.	1
10	CAlternate angles between parallel lines are equal, so the alternate angle is 78°.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
1(a)	180° - 68° = 112°, because co-interior angles between parallel lines are supplementary (sum to 180°). 112° [1] Reason: co-interior angles sum to 180° [1]	2
1(b)	68°, because alternate angles between parallel lines are equal. 68° with the alternate-angles reason [1]	1
1(c)	The angle measures 68°. [1]	1
1(d)	The alternate angle, the vertically opposite angle, and the corresponding angle (on the same side of the transversal at the other parallel line) all equal 68°. Any two of alternate / vertically opposite / corresponding named [1]	1
	Total for Question 1	5
2(a)	(x + 20) + (2x + 10) + 60 = 180 Uses the angle sum of a triangle, 180° [1] Correct equation [1]	2
2(b)	3x + 90 = 180, so 3x = 90 and x = 30. Collects to 3x = 90 [1] x = 30 [1]	2
2(c)	The angles are 50°, 70° and 60°, so the largest is 70°. [1]	1
	Total for Question 2	5
3(a)	360° ÷ 5 = 72° Uses the exterior angle sum 360° [1] 72° [1]	2
3(b)	180° - 72° = 108° [1]	1
3(c)	The number of sides would be 360 ÷ 50 = 7.2, which is not a whole number. A polygon must have a whole number of sides, so no regular polygon has exterior angle 50°. Uses number of sides = 360 ÷ exterior angle [1] Shows the result 7.2 is not a whole number [1]	2
	Total for Question 3	5
4(a)	Base angles are equal, and they sum with 36° to 180°: (180 - 36) ÷ 2 = 72° each. Uses the isosceles property (equal base angles) [1] 72° each [1]	2
4(b)	Angle ACD = 180° - 72° = 108° (angles on a straight line). Angles on a straight line used [1] 108° [1] Accept the exterior-angle shortcut: 36° + 72° = 108°.	2
4(c)	The exterior angle equals the sum of the two opposite interior angles: 36° + 72° = 108°, which agrees with the answer in (b). States exterior angle = sum of opposite interior angles with the numbers [1]	1
	Total for Question 4	5

— End of mark scheme —