Chapter 7 Practice Paper: Pythagoras’ Theorem and Irrational Numbers — mark scheme (print)

QuizQuipPractice Series

QQ/S2MATH/CH7

Secondary 2 · Lower Secondary Mathematics (G2 / G3)

Mark scheme

Pythagoras’ Theorem and Irrational Numbers

Chapter 7 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	C81 + 144 = 225, and √225 = 15.	1
2	A289 − 64 = 225, and √225 = 15.	1
3	DThe others are fractions of whole numbers; √7 is not.	1
4	B25 + 49 = 74, which is not a perfect square, so the exact form is √74.	1
5	B49 + 576 = 625 = 25², so (7, 24, 25) passes the converse test; the other triples fail it.	1
6	A36 < 40 < 49, so 6 < √40 < 7.	1
7	Dd² = 64 + 64 = 128, so d = √128 = 8√2.	1
8	B100 − 36 = 64, and √64 = 8 m.	1
9	CThe converse turns the formula into a right-angle test.	1
10	AWith no repeating block, no fraction can produce this decimal.	1

Section B: Structured questions[20 marks]

Question	Answer and marking points	Marks
1(a)	h² = 81 + 144 = 225, so h = 15. 81 + 144 = 225 [1] 15 [1]	2
1(b)	9 + 12 + 15 = 36 [1]	1
1(c)	h² = 400 + 441 = 841, so h = 29. 400 + 441 = 841 [1] 29 [1]	2
	Total for Question 1	5
2(a)	The converse of Pythagoras’ theorem Converse named [1]	1
2(b)	11² + 60² = 121 + 3600 = 3721 = 61², so the angle opposite 61 is a right angle. 121 + 3600 = 3721 computed [1] Equals 61², so right-angled [1]	2
2(c)	61 (the longest side, opposite the right angle) [1]	1
2(d)	No — all three sides have different lengths. No, sides all differ [1]	1
	Total for Question 2	5
3(a)	w² = 2500 − 900 = 1600, so w = 40 inches. 2500 − 900 = 1600 [1] 40 inches [1]	2
3(b)	40 × 30 = 1200 square inches 1200 in² [1]	1
3(c)	d² = 2304 + 400 = 2704, so d = 52 inches. 2304 + 400 = 2704 [1] 52 inches [1]	2
	Total for Question 3	5
4(a)	x² + 81 = 225. x² + 9² = 15² formed [1]	2
4(b)	x² = 144, so x = 12 (or −12 algebraically). x² = 144 [1] x = 12 noted [1]	2
4(c)	x is a length, and lengths are positive, so only 12 is valid. Lengths are positive [1]	1
	Total for Question 4	5

— End of mark scheme —