Mathematics Success Guide & Exercises (Secondary 2)

Tips and reminders for problem-solving

Enhances thorough understanding to the question

Common mistakes among students

Special points students need to pay attention to

Avoids students from repeating the same mistakes

Reveals the key point to answer the question

Leads students to the right track for problem-solving

Students used to adopt the fastest possible method in

problem-solving during exams. “Alternatives” provides

you with a variety of methods to solve a mathematical problem.

“Easy as Pie”

The questions are standard and straightforward. Students are

supposed to handle these questions fast with accuracy.

Solution Part

Exercise Part

“Go Banana”

The questions are complicated or tricky. Students may have to

spend more time and pay more effort to go through.

“DSE Zone”

The questions are designed for preparing DSE.

Contents

1 Errors in Measurement

Key Points 7 Exercise – Multiple Choice 9 Exercise – Short and Long Questions 12

2 Rate and Ratio


3 Index


4 Polynomials


5 Identities


6 Algebraic Fractions

Key Points 77 Exercise – Multiple Choice 79 Exercise – Short and Long Question 83

7 Change of Subjects


SUGGESTED SOLUTIONS are provided after the last session.

Contents

8 Equations in Two unknowns


9 Basic Geometry


10 Pythagoras’ Theorem


11 Trigonometric Ratios


12 Area and Volume


13 Data Handling


14 Revision Test Exercise – Multiple Choice 182 Exercise – Short and Long Questions 186

CHAPTER10PYTHAGORAS’THEOREM

135

10 Pythagoras’ Theorem

Key Points 10.1 Square roots

e.g. e.g. Square roots of 36 are 6 and –6. (6 is the positive square root and –6 is the negative square root) e.g. Solve x2 = 36

x = x = 10.2 Rational and irrational numbers

Definition of Rational Number

A rational number is any number that can be expressed in the form of (where a is an

integer and b is a non-zero integer). Examples of rational numbers: Positive integers/ zero / negative integers (e.g. 2, 0, –6);

Proper fractions/ improper fractions/ mixed fractions (e.g. , , );

Terminating decimals (e.g. 1.5) and recurring decimals (e.g. ) Examples of irrational numbers: Non-terminating decimals (e.g. 𝜋, , 1.41237…) 10.3 Pythagoras’ theorem

In ABC, If = 90o, then AB 2 + BC 2 = AC 2. (Pyth. thm.)

636 =

36±6±

ba

2138

321

•

3.0

6

D

ABCÐ

A

B C

hypotenuse


136

10.4 Converse of Pythagoras’ theorem In ABC, If AB 2 + BC 2 = AC 2, then = 90o. (Converse of Pyth. thm.) e.g. Determine whether PQR is a right-angled triangle. Sol: PR2 + PQ2 = 62 + 82 = 100 = QR2 (Converse of Pyth. thm.)

PQR is a right-angled triangle.

Remarks l “ ” is called “radical sign”.

l Hypothenuse is the longest side of a triangle.

Relevant Vocabularies Square root 平方根 Rational number 有理數 Irrational number 無理數 Integer 整數 Proper fraction 真分數 Improper fraction 假分數 Mixed fraction 帶分數 Terminating decimal

有盡小數 Non-terminating decimal 無盡小數

Recurring decimal 循環小數

Pythagoras’ theorem 畢氏定理

Converse of Pythagoras’ theorem畢氏定理的逆定理

Hypotenuse 斜邊

D

ABCÐ

Δ

∴∠QPR = 90°∴Δ

A

B C

hypotenuse

P

Q

R

6 8

10


137

Score: 18

Exercise – Multiple Choice There are 9 questions. Each carries 2 marks. 1. In the figure, is a right-angled isosceles triangle. If BC = 12,

the area of the triangle is

Ans

A. 18 sq. units B. 24 sq. units C. 36 sq. units D. 48 sq. units

2. In the figure, is a right-angled triangle. If DC = CB, then

AC =

Ans

A. 6.47 units B. 6.90 units C. 7.21 units D. 7.47 units

3.

In the figure, the height of trapezium ABCD is 4. If AB = 8, AD = 6, BC = 10, then the area of trapezium is

Ans

A. 58.0 sq. units B. 59.3 sq. units C. 68.0 sq. units D. 69.3 sq. units

4. In the figure, ABCD is a rectangle. AC is the diagonal while AC BE.

If AD = 18 and DC = 24, then BE =

Ans


ABCD

ABDD

^

A

B C

A

B C D

6A

10

A

C

B

D

8

6 10

`

B A

C D

E 18

24


138

5. Which of the following is NOT an irrational number?

Ans

A.

B.

C. 3.14 D. 22 7

6. The diagonal of a rectangle is 30 cm. If the ratio of the length and the

width is 4 : 3, then the area of the rectangle is

Ans

A. 150 cm2 B. 210 cm2 C. 360 cm2 D. 432 cm2

7. If ABC is a right-angled isosceles triangle and AC is the

hypothenuse, then AB =

Ans

A.

B.

C.

D.

8. In the figure, AB = CD = 3, BC = 12, ED = EF = 2, find the distance

between A and F.

Ans


9. In the figure, AEB and ADC are straight lines. It is given that AE = 7,

EB = CB = 8, find ED.

Ans

A. 4.41 B. 4.57 C. 5.23 D. 6.34

22× 7π22

π × 4π 2

÷π π ÷ ×

D

AC21

AC22

AC23 AC2

B

A

C

D E

F

A

B C

D

E

8

8

7


139

Score: 39

Exercise – Short and Long Questions

1. In the figure, AB = 3.2, AD = and CD = 0.8. Determine whether the value of BC is a rational number.

(3 marks)

2. In the figure, AED and BCF are right-angled isosceles triangles and

ABCD is a square. It is given that the AC = 20 cm, find the area of the figure. (5 marks)

3.84

D D

A B

C D

E

F

A

B C

D


142

5. In the figure, . It is given that AB = 18, ED = 14,

BD = 20, BC = 12 and DC = 16, find the area of the figure, correct your answer to 3 significant figures.

(5 marks)

°=Ð=Ð 90AEDBAE

A B

C

D E

20

16

12

18

14


144

7. In the figure, ABCD is a square. F and G are points lying on BC and DC

respectively. DF and AG intersect at E. It is given that AE = 16, EG = 9 and DG = 15.

(a) Is that a right-angled triangle? Explain your answer. (5 marks)

(b) Find DE and EF. (6 marks)

DEGD

A

F

B

C D G

E

16

9

15

CHAPTER10–PYTHAGORAS’THEOREMSOLUTIONS

90

Solutions Multiple Choice 1. C 2. C 3. B 4. B 5. B 6. D 7. B 8. A 9. A 1. C

AB 2 + AC 2 = BC 2 (Pyth. thm) 2AB 2 = 144 AB 2 = 72 The area of the triangle

=

= 36 sq. units

2. C DB = (Pyth. thm) DC = 4

AC = (Pyth. thm)(corr. to 3 sig. fig.)

3. B

DE = (Pyth. thm)

FC = (Pyth. thm) Area of the trapezium

=

= 59.3 sq. units (corr. to 3 sig. fig.)

2

2AB

864601 22 ==-

42 + 62 = 52 = 7.21

2046 22 =-

84410 22 =-

482484

2420

´+´

+´

Idea: As the hypotenuse is the longest side in the triangle, therefore this isosceles triangle must refer to “AB = AC ”.

A

C

B

D

8

6 10 4

F E

Graphical idea:


91

4. B AC = (Pyth. thm) area of = area of

BE = 14.4 units

5. B Using the calculator, answers A, C and D give non-recursive numbers. Only answer B gives the value of 2, which is obviously a rational number.

Alternatively,

6. D Let the length and the width of the rectangle be 4k and 3k, (4k)2 + (3k)2 = 302 (Pyth. thm) 16k2 + 9k2 = 900 25k2 = 900 k2 = 36 The required area = (4k)(3k) = 12k2 = 12(36) = 432 cm2

7. B AB 2 + BC 2 = AC 2 (Pyth. thm) 2AB 2 = AC 2

AB =

309002418 22 ==+ADCD ABCD

230

22418 BE´

=´

π × 4π 2 = π × 4

π 2

= π × 2π= 2

2

2AC

AC21

=

Idea: As the hypotenuse is the longest side in the triangle, therefore isosceles triangle must refer to “AB = BC ”.


92

8. A AG = AB + CD – EF = 4 GF = BC – DE = 10

AF = units (corr. to 3 sig. fig.)

9. A Consider ADE and ABC,

(given) DAE = BAC (common ) ADE = ABC ( sum of ) ADE ~ ABC (AAA)

Note that AC 2 = AB 2 – CB 2 (Pyth. thm)

AC =

(corr. sides, ~ )

DE = 4.41 (corr. to 3 sig. fig.)

AC22

21´=

AC22

=

8.10116104 22 ==+

D D°=Ð=Ð 90ACBAED

Ð Ð ÐÐ Ð Ð DD D

(7 +8)2 −82 = 161

ACAE

BCDE

= D

1617

8=

DE

Graphical idea:

Idea: The question is obviously related to similar triangles.

B

A

C

D E

F G


93

Solutions

Short and Long Questions 1. In ,

AC = (pyth. thm.)

In ,

BC = (pyth. thm.)

which is a rational number.

1M 1A 1A

2 AB2 + BC2 = 202 (Pyth. thm) 2AB2 = 400 AB2 = 200 Consider AED, AE2 + ED2 = AD2 (Pyth. thm) 2ED2 = 200 ED2 = 100

Area of AED = cm2

Area of BCF = cm2

The total area = 200 + 50 + 100 = 350 cm2

1M 1A 1M 1A 1A

3. Let DB = x, In DBC, BC 2 = 202 – x2 (Pyth. thm) In ABC, BC 2 = 342 – (x + 18)2 (Pyth. thm) So, 202 – x2 = 342 – (x + 18)2 400 – x2 = 1156 – (x2 + 36x + 324) 36x = 432 x = 12 Therefore,

BC = (Pyth. thm)

1M 1A 1M 1A 1A

ΔACD

( 3.84)2 + 0.82 = 4.48

ΔABC

3.22 − ( 4.48)2 = 5.76 = 2.4

∵BC = 2.4 = 125

D

D 502100

2

2

==ED

D 1002200

2

2

==BC

D

D

161220 22 =-

Reminder: (a + b)2 = a2 + 2ab + b2


94

4. Refer to the figure, BG = AE = 12 In ,

DG = (Pyth. thm)

CG = In ,

BC = units (Pyth. thm) (corr. to 3 sig. fig.) AD = BC = 13.4 units

1M+ 1A 1M+ 1A

5. Consider that BD 2 + DC 2 = 122 + 162 = 400 = BD 2 So, (converse of Pyth. thm) FB = 18 – 14 = 4 In FBD,

FD = (Pyth. thm) Therefore, the required area

=

= 410 sq. units

1M 1A 1M +1A 1A

BDGD

11521236 22 =-

281152 -BCGD

BG2 +CG2 = 13.4

°=Ð 90BCD

D

384420 22 =-

21612

2384)1418( ´

+´+

Graphical idea:

E D C

B A

F

G

A B

C

D E

20

16

12

F

14

Graphical idea:


95

6. (a) AC2 + CB2 = 122 + 162 = 400 = AB2 So, (converse of Pyth. thm) and ACD = 90o (adj. s on st. line) Let ABC = x, then CAB = 90o – x ( sum of ) then DAC = 90o – (90o – x) = x then ADC = 90o – x ( sum of ) Therefore, (AAA)

1M 1A 1M 1A

(b) (corr. sides, ~ )

AD = 15

the area = sq.units

1M 1A 1A

7. (a) Consider and ,

DGE = AGD (common )

and

So,

~ (ratio of 2 sides, inc. ) DEG = ADG = 90o (corr. s, ~ )

Therefore, is a right-angled triangle.

1M 1M+ 1A 1M 1A

(b) DE = (Pyth. thm) EDG = DAG (corr. s, ~ ) ADG = DCF and AD = DC (properties of square)

≅ (ASA) DF = AG = 25 (corr. sides,≅ ) EF = 25 – 12 = 13

1M+ 1A 1M 1M+ 1A 1A

°=Ð 90ACBÐ ÐÐ Ð Ð DÐÐ Ð D

BACADC DD ~

BCAC

BAAD

= D

1612

20=

AD

15022015

=´

ADGD DEGDÐ Ð Ð

35

1525

==DGAG

35

915

==EGDG

EGDG

DGAG

=

ADGD DEGD ÐÐ Ð Ð D

DEGD

12915 22 =-Ð Ð Ð DÐ ÐADGD ΔDCF

D

Idea: Similar triangles again!

Mathematics Success Guide & Exercises (Secondary 2)

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Transcript of Mathematics Success Guide & Exercises (Secondary 2)