Secondary 2 Workbook Answer Key Chapter 1 Real Numbers and ...€¦ · Page 1 of 37 Secondary 2...

Page 1 of 37

Secondary 2 Workbook Answer Key

Chapter 1 Real Numbers and Its Operations

1.1 Square Roots

A. Definition of Square Roots

1. C

2.（1）Yes （2）Yes （3）No （4）Yes

Reasoning: The numbers under the square root must be non-negative.

3.（1） 14 （2）5

3 （3）

3

1

10 （4） 0.04

4. 4

5. （1） 1.2 （2） 9 （3）7

2 （4）

6

5

6. 49

B. Definition of Principal Square Root

1. （1）1.1 （2）0.2 （3）5

6 （4）10

2. 0 or 1 ， 0

3. 0.4，6

5，3，0.5

4. -1 or 7

C. Application of Principal Square Root

1. 3 ， - 2 2. < ， >

3. 9 4. 1 5. -6

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1.2 Cube Roots

A. Definition of Cube Roots

1. A 2. B 3. A

4.（1） -0.3 （2）2

-3

（3）3

4 （4）

1-

2

5. （1）3

-2

（2）1

2 6. （1）5 （2）

7-

4

B. Application to Cube Roots

1. B 2. （1）1 （2）5

3

3. 1𝑐𝑚 4. 2， 3， 3 n 5. 4

1.3 Understanding Real Numbers Again

A. Understanding 4th Root and nth Root

1. （1） 1 （2） 3 （3）Has no real solution. （4）0

2. 3

2 3. - 1 0

B. Real Numbers and Number Line

1. C 2. B 3. D 4. B 5. C 6. D 7. 2 3

1.4 Operations on Quadratic Radicals

1.4.1 Definition of Quadratic Radicals

1. A 2. D 3. C

4.（1） 3- 2 （2）4-2x （3）𝜋 − 3 （4）1

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5. 2 6. 1 7. 2+ 3-2

1.4.2 Multiplying and Dividing Quadratic Radicals

A. Multiplying Quadratic Radicals

1. C 2. -4 3a

3.（1） 24 2x （2） 2 5 b ab （3）2 3b

4.（1）8 6 （2）5 5

2 （3） 24 a b

（4）20 （5）2 30

25 （6）3 3 2

5. 2 11 3 5 4 3

B. Dividing Quadratic Radicals

1. 2 2a 2.（1）5

3 （2）

3 2

4 （3）

2

x

3.（1）2 （2）2 3 （3）1 （4）1

9 （5）

3 3

2 （6）45 6

C. Simplifying Quadratic Radicals

1. C

2.（1）3

6 （2）

30

10 （3）

10

5 （4）

15

5

ab

a （5）

2

x x （6）

10 5

5

3. 4x xy . The value is 25

3.

1.4.3 Adding and Subtracting Quadratic Radicals

1.（1）3𝑎 （2） 3 2 （3）2 13 3

4 3

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（4）6 3 5 （5）13 3 （6）2 3

1.4.4 Mixed Operations on Quadratic Radicals

1.（1）8 7 （2）2 3 1 （3）6 10 6 （4）9

（5）2 （6）0 （7） 32

ab （8）

7 23

2

2. 2

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Quiz Yourself (A)

I. Multiple choice questions.

1. A 2. C 3. D 4. C 5. D 6. A 7. A 8. C 9. C

II. Fill in the blanks.

10. 4 ， 6 ，2，2

5 11. 0.5，

1

3，

1

3 ，18

12. 2 13. 3a 14. 2, 1,0,1,2 15. <

III. Short answer questions.

16.（1）7 （2）2

3 （3）

2

3 （4） 23a

（5） 1 （6） 2 （7） 3 2

17. 81

Quiz Yourself (B)


1. A 2. C 3. C 4. A 5. B 6. B


7. 1

2x 8. a 9. 3, 1 2x x 10. 2x 11. 1


12. 1

2a . The minimized value is 1.

13.（1）2

b

ab （2）2 3 （3）60 3 （4） 3 （5）

9 2

8 （6）3 x

14. 1

Page 6 of 37

Chapter 2 Polynomial Operations

2.1 Adding and Subtracting Polynomials

A. Definition of Polynomials

1. ③④⑤，①②⑥，①②③④⑤⑥

2. Degree: 2. Terms: 3. The 1st term: −2𝑥. The 2nd term: −5. The 3rd

term: 3

3. Answer not provided.

4.（1） 8 92 x y （2） 1 1( 1) 2n n nx y

B. Like Terms

1. B 2. D 3. 30𝑥, 60 4. 2，3，1

5. 3

C. Combining Like Terms

1.（1）3ab （2） 2 23 20 17ab a b （3）5

3y （4） 214 10

23 3

ab a

2. 8 3. 11

D. Removing and Inserting Parenthesis

1.（1） 12 6x （2） 5 x

2. 3 26 1x x x 3. 3 2 1x x 4. 1

2b a 5. 0

6.（1） 2 22 2x y （2） 2 218 22x y （3） 2 23 4a b ab

（4）5 1y （5） 2 23 4a b ab （6） 25 2 3x x 7. 7

E. Adding and Subtracting Polynomials

1.（1） 14 （2） 6

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2. 22 7 3x x

3. 𝐴 − 𝐵 = 21 54

6 2x y xy ，2𝐴 − 3𝐵 = 2 13

2 112

x y xy

2.2 Multiplying Polynomials

A. Multiplying Monomials

1.（1） 9 6125x y z （2） 4 8 312 16x y x y （3） 3 26x y （4） 6 334a b

2. 22

B. Multiplying with Monomial and Polynomial—Multiplication

Involving 𝒂(𝒎 + 𝒏 + 𝒑)

1.（1） 3 28 12 4a a a （2） 2 3 2 21

3a b a b （3） 4 2 3 3 2 420 8 12x y x y x y

（4） 4 5 3 5 2 712 8 16a b a b a b （5） 3 23 4 14x x x

2. 36 3. 19

34

C. Multiplication with Polynomial and Polynomial—Multiplication

Involving (𝒂 + 𝒃)(𝒎 + 𝒏)

1.（1） 2 2 35x x （2） 23 5 2x x （3） 2 24 9m n

（4） 2 24 +12 9a ab b （5） 2 2x y x y

2. 1 2x 3. 0x 4. 222 24x x

2.3 Multiplication Formulas

A. Square of Sums or Difference of Two Numbers—Multiplication

Involving (𝒂 ± 𝒃)𝟐

Page 8 of 37

1. (1) 9𝑚2 − 24𝑚𝑛 + 16𝑛2 (2) 𝑚2 +2

3𝑚𝑛 +

1

9𝑛2

(3) 4

9𝑥2 − 2𝑥𝑦 +

9

4𝑦2 (4) 32𝑎2 = 2𝑏2

(5) −33𝑥2 + 33 (6) 7921

9

2. (𝑎 + 𝑏)2 and (−𝑎 − 𝑏)2 are equal because (−𝑎 − 𝑏)2 = [−(𝑎 +

𝑏)]2 = (𝑎 + 𝑏)2.

(𝑎 − 𝑏)2 and (𝑏 − 𝑎)2 are equal because (𝑏 − 𝑎)2 = [−(𝑎 − 𝑏)]2 =

(𝑎 − 𝑏)2.

(𝑎 − 𝑏)2 and 𝑎2 − 𝑏2 are not equal because (𝑎 − 𝑏)2 = (𝑎 − 𝑏)(𝑎 −

𝑏) while 𝑎2 − 𝑏2 == (𝑎 + 𝑏)(𝑎 − 𝑏).

3. 𝑎2 + 𝑏2 =5

2, 𝑎𝑏 =

1

4.

4. 9

2

B. Multiplication with sum and Difference of two Numbers—

Multiplication Involving (𝒂 + 𝒃)(𝒂 − 𝒃)

1. A 2. B

3.（1） 2 249 9y x （2） 2 216x y （3） 2 125

16x （4） 2 21 1

9 4m n

4.（1）9996 （2）24

39925

5. （1） 220 20m （2） 42 16x

2.4 Dividing Polynomials

2.4.1 Dividing Monomials

1.（1） 2 22x y （2） 11ac （3） 25m （4） 3 36a b （5） 26 ( )a m n

2.（1） 3a （2） 3 33( )

2x a b 3. 2, 1m n 4. 4x

Page 9 of 37

2.4.2 Dividing a Polynomial by a Monomial

1.（1） 5 a （2） 3 2 34

2x x （3）

22

7a b

2. （1） 3 44

3x x （2） 2 28 6 12xy x y （3）

14

2x

3. 4

2.4.3 Dividing a Polynomial by a Polynomial

1. 2 2 1x x 2. 2 3x x 3. 1

2k 4. 1

2.5 Factoring

2.5.1 Factoring by Common Factor—Factoring Involving 𝒂𝒎 + 𝒂𝒏 +

𝒂𝒑

1. ( )m a b c ， 2 (1 2 )x y ，2 ( 2 1)xy x y ， 3( 2 )m m n n

2.（1） 2 22 (2 )x x y （2） (3 6 1)x xy y （3） 2 24 2 3 )ab a bc（

（4） ( 2 )( 2)x y x （5） ( 2)(2 3 )a x y （6） 22(1 ) ( 1)p q pq

（7） )(2 3x y xy （）（8） (2 )( 3 )a b a b

3. 1999

2.5.2 Difference of Squares—Factoring Involving 𝒂𝟐 − 𝒃𝟐

（1） 2(2 2 1)x y （2） xy z xy z ( ) ( ) （3）1 2 1 2

3 11 3 11x y x y ( ) ( )

（4） 2 2mn m n m n 2 ( ) ( ) （5） 0.6 -0.6xy xy( ) ( ) （6）1 1

2 -2 2

m m( ) ( )

（7）a a b( 2 ) （8） 5x y x y - ( 5 ) ( )

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2.5.3 Perfect Square Trinomials—Factoring Involving 𝒂𝟐 ± 𝟐𝒂𝒃 + 𝒃𝟐

1. A 2. 2

3.（1） 2(5 )m n （2） 2(4 5 )m n （3） 22( )3

x y （4） 2(2 2 1)x y

（5） 2(2 2 3)x y （6） 2(2 3)m （7） 2(2 3 )m n （8） 2(7 1)a

2.5.4 Introduction to Factoring Higher Order Polynomials

（1） 2 22 ( 2 )y x yz （2） 2( 2)x x （3） 33 ( 3)( 3)x x x

（4） 2 2( 2 )( 2 2 )x x y x xy y （5） 2 2(2 1 2 1)a a ）（（6） 2( 2)( 2)( 2)x x x

（7） 2 23 ( )( )a x y x y （8） 2 2(4 25 )(2 5 )(2 5 )m n m n m n

（9） 2 2( +1) ( 2 1)p p p （10） 2 2( +2) ( 2)x x

2.6 Applications to Polynomial Operations

1. 220 4a ab 2. Answer not provided.

3. （1） 2(2 1)(2 1 = 2 ) 1n n n ）（（2）Answer not provided.

Page 11 of 37

Quiz Yourself (A)


1. B 2. D 3. B 4. D 5. D 6. C

7. D 8. D 9. C 10. C 11. D


12. 86a , 7ab , 64a 13. 2 6x x 14. 29 12 4x x

15. (9 )(9 )x x 16. 6 17. 2 2b c 18. 80 19. 3，1


20.（1） 2 210a ab b （2）3 2x

21. （1） 14n （2）1

2ac （3） 2 29 4a b （4） 4 412a b （5） 24 10xy y

（6） 4 41 3 27

5 5 25x y x y

22. (1) 3𝑎(𝑎 − 3𝑏) (2) (2𝑚 + 2𝑛)(3𝑚 − 2𝑛)

(3) 2𝑎(𝑎 − 3𝑏)2

23. 3

2

24.（1）34 （2）15

Quiz Yourself (B)


1. B 2. A 3. A 4. B 5. A 6. A


7. 2 8. 4037 9. 4 10. 2

Page 12 of 37


11（1）0 （2）37

16

12.（1）25 （2）25. The two answers are equal.

13.（1）Yes. 2 228 8 6 ，2 22012 504 502

(2) Answer not provided.

(3) Answer not provided.

Page 13 of 37

Chapter 3 Geometry

3.1 Plane and Solid

1．D

2.

Cylinder Hexagonal prism Cone Triangular pyramid

Cube Cuboid Rectangular pyramid Triangular prism

3．（1）Point 1 （2） Point 3 or point 4 （3）Point 1 or point 6

4.

The distance between A and C is the shortest, because the line segment

Page 14 of 37

gives the shortest distance between two points.

3.2 Surface Area of a Solid

1. B 2. 4 3. 126 4. 96 5. 3 6. 𝑙 = 6𝑐𝑚.

3.3 Volume of Solid

1. 96π 2. 2.5 3. 75 4. 3 10 5.1000

6. 2

9𝜋 7. 2 8. 4𝑎 9.

25

12

3.4 Congruence of Triangles

3.4.1 Recognizing Congruence Triangles and Their Properties

1．B 2．C 3．C ①③④ are correct. 4．D 5．150°

3.4.2 Criteria of Congruence Triangles

A. Side-Side-Side (SSS) Congruence and Compass and Straightedge

Construction on an Angle That Equals to a Given Angle

1. Proof. ∵ 𝐴𝐵 = 𝐷𝐶, 𝐴𝐶 = 𝐷𝐵, 𝐵𝐶 = 𝐵𝐶

∴ △ 𝐴𝐵𝐶 ≌△ 𝐷𝐶𝐵 (𝑆𝑆𝑆)

2. Proof. ∵ 𝐴𝐶 = 𝐵, 𝐵𝐶 = 𝐴𝐷，𝐴𝐵 = 𝐴𝐵

∴ △ 𝐴𝐵𝐶 ≌△ 𝐵𝐴𝐷 (𝑆𝑆𝑆)

∴ DC

B. Side-Angle-Side (SAS) Congruence

1. Proof. ∵ 𝐴𝐷 = 𝐵𝐶, ∠𝐴𝐷𝐶 = ∠𝐵𝐶𝐷, 𝐷𝐶 = 𝐷𝐶

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∴ △ 𝐴𝐷𝐶 ≌ 𝐵𝐶𝐷 (𝑆𝐴𝑆)

∴ ∠𝐴𝐷𝐶 = ∠𝐵𝐶𝐷, ∠𝐵𝐷𝐶 = ∠𝐴𝐶𝐷

∴ ∠𝐴𝐷𝐶 − ∠𝐵𝐷𝐶 = ∠𝐵𝐶𝐷 − ∠𝐴𝐶𝐷

∴ ∠𝐴𝐷𝐵 = ∠𝐵𝐶𝐴.

2. Proof. ∵ 𝐴𝐵 = 𝐴𝐶, ∠𝐴 = ∠𝐴, 𝐴𝐷 = 𝐴𝐸

∴ △ 𝐴𝐵𝐸 ≌△ 𝐴𝐶𝐷 (𝑆𝐴𝑆)

∴ 𝐶𝐷 = 𝐵𝐸

C. Angle-Side-Angle (ASA) and Angle-Angle-Side (AAS) Congruence

1. Proof. ∵ AD bisects ∠𝐵𝐴𝐶, ∴ ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷

∵ ∠𝐵 = ∠𝐶，∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷，𝐴𝐷 = 𝐴𝐷

∴ △ 𝐴𝐷𝐵 ≌△ 𝐴𝐷𝐶 (𝐴𝐴𝑆)

∴ ∠𝐴𝐷𝐵 = ∠𝐴𝐷𝐶

Also, ∵ ∠𝐴𝐷𝐵 + ∠𝐴𝐷𝐶 = 180°

∴ ∠𝐴𝐷𝐵 = ∠𝐴𝐷𝐶 = 90°

∴ 𝐴𝐷 ⊥ 𝐵𝐶

2．Proof. ∵ 𝐴𝐵 ∥ 𝐷𝐶, 𝐴𝐷 ∥ 𝐵𝐶

∴ ∠1 = ∠2, ∠3 = ∠4

∵ ∠1 = ∠2, 𝐵𝐷 = 𝐷𝐵, ∠3 = ∠4

∴ △ 𝐴𝐵𝐷 ≌△ 𝐶𝐷𝐵 (𝐴𝑆𝐴)

∴ 𝐴𝐵 = 𝐷𝐶

3. Proof. ∵ ∠1 = ∠2

∴ ∠1 + ∠𝐷𝐴𝐶 = ∠2 + ∠𝐷𝐴𝐶

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∴ ∠𝐵𝐴𝐶 = ∠𝐷𝐴𝐸

∵ ∠𝐵 = ∠𝐷, 𝐴𝐵 = 𝐴𝐷, ∠𝐵𝐴𝐶 = ∠𝐷𝐴𝐸

∴ △ 𝐴𝐵𝐶 ≌△ 𝐴𝐷𝐸 (𝐴𝑆𝐴)

∴ 𝐵𝐶 = 𝐷𝐸

D. Hypotenuse-Leg (HL) Congruence

1. Proof.（1）∵ 𝐷𝐸 ⊥ 𝐴𝐶, 𝐵𝐹 ⊥ 𝐴𝐶

∴ ∠𝐷𝐸𝐶 = ∠𝐵𝐹𝐴 = 90°

∵ 𝐴𝐵 = 𝐶𝐷，𝐷𝐸 = 𝐵𝐹.

∴ △ 𝐷𝐸𝐶 ≌△ 𝐵𝐹𝐴 (𝐻𝐿)

∴ 𝐴𝐹 = 𝐸𝐶

∴ 𝐴𝐹 − 𝐸𝐹 = 𝐸𝐶 − 𝐸𝐹

∴ 𝐴𝐸 = 𝐶𝐹

（2）∵ △ 𝐷𝐸𝐶 ≌△ 𝐵𝐹𝐴

∴ ∠𝐴 = ∠𝐶

∴ 𝐴𝐵 ∥ 𝐶𝐷

2. Proof. ∵ AD is the height of △ 𝐴𝐵𝐶.

∴ ∠𝐵𝐷𝐹 = ∠𝐴𝐷𝐶 = 90°

∵ 𝐵𝐹 = 𝐴𝐶, 𝐹𝐷 = 𝐶𝐷

∴ △ 𝐵𝐷𝐹 ≌△ 𝐴𝐷𝐶 (𝐻𝐿)

∴ ∠𝐷𝐵𝐹 = ∠𝐷𝐴𝐶

∵ ∠𝐷𝐴𝐶 + ∠𝐶 = 90°

∴ ∠𝐷𝐵𝐹 + ∠𝐶 = 90°

http://www.czsx.com.cn/

Page 17 of 37

∴ ∠𝐵𝐸𝐶 = 90°

∴ 𝐵𝐸 ⊥ 𝐴𝐶

3.4.3 Angle Bisector

1. C 2. A 3. C 4. B 5. 2cm 6. 3

7. 𝐵𝐶 = 𝐴𝐵 + 𝐴𝐸

(Hint for the proof. First construct 𝐸𝐹 ⊥ 𝐵𝐶. Then prove that triangle EFC

is a right isosceles triangle to obtain 𝐸𝐹 = 𝐹𝐶.

Since BE bisects ∠𝐴𝐵𝐶, we can have

𝐴𝐸 = 𝐸𝐹 = 𝐹𝐶. Then prove that

△ 𝐴𝐵𝐸 ≌△ 𝐹𝐵𝐸 to obtain 𝐴𝐵 = 𝐵𝐹,

so then 𝐵𝐶 = 𝐵𝐹 + 𝐹𝐶 = 𝐴𝐵 + 𝐴𝐸. )

3.4.4 Perpendicular Bisector

1. 8 2. 80° 3.（1）80° （2）50cm．

4． 5. 20

3.4.5 Reflection Transformation

1． B 2． C 3. D 4．0

Page 18 of 37

5．

A1(2, 3), B1(3, 2), C1(1,1)

6.（1）（2）

（3）

3.5 Isosceles Triangles

3.5.1 Properties of Isosceles Triangles

1. B 2. D 3. 50°, 50° or 20°, 80°

4. Hint for the proof.

First prove that △ 𝑂𝐷𝐵 ≌△ 𝑂𝐸𝐶 (𝐴𝐴𝑆) to obtain 𝐵𝐷 = 𝐶𝐸. Then from

𝐴𝐵 = 𝐴𝐶 to obtain 𝐴𝐷 = 𝐴𝐸.

Page 19 of 37

5．（1）Since 𝐴𝐵 = 𝐴𝐶, and point D is the midpoint of BC, then we

have ∠𝐵𝐴𝐷 = ∠𝐶𝐴𝐷. Using SAS to obtain △ 𝐵𝐴𝐷 ≌△ 𝐶𝐴𝐷.

（2） Since △ 𝐵𝐴𝐷 ≌△ 𝐶𝐴𝐷, we know AD perpendicularly bisects BC,

then we have 𝐵𝐸 = 𝐶𝐸.

6．Yes, the statement 𝐷𝐸 = 𝐷𝐵 + 𝐸𝐶 is true.

From 𝐷𝐸 ∥ 𝐵𝐶 and the angle bisectors of ∠𝐴𝐵𝐶 and ∠𝐴𝐶𝐵 to obtain

𝐵𝐷 = 𝐷𝐹, 𝐶𝐸 = 𝐸𝐹. Then we can have 𝐷𝐸 = 𝐷𝐹 + 𝐸𝐹 = 𝐷𝐵 + 𝐸𝐶.

3.5.2 Criteria for Isosceles Triangles

1. B 2. C 3. C 4. B 5. A

6. 3 7. 5 8. 72°, 3 9. Isosceles

10．Proof. 𝑂𝐸 ⊥ 𝐴𝐵 at E, 𝑂𝐹 ⊥ 𝐴𝐶 at F.

∵ AO bisects ∠𝐵𝐴𝐶.

∴ 𝑂𝐸 = 𝑂𝐹.

∵ ∠1 = ∠2

∴ 𝑂𝐵 = 𝑂𝐶.

∴ 𝑅𝑡 △ 𝑂𝐵𝐸 ≅ 𝑅𝑡 △ 𝑂𝐶𝐹(𝐻𝐿).

∴ ∠5 = ∠6．

∴ ∠1 + ∠5 = ∠2 + ∠6．

That is, ∠𝐴𝐵𝐶 = ∠𝐴𝐶𝐵．

∴ 𝐴𝐵 = 𝐴𝐶．

∴ △ABC is an isosceles triangle.

Page 20 of 37

11．Solution: In △ 𝐴𝐵𝐶,

∵ 𝐴𝐵 = 𝐴𝐶，∠𝐴 = 36°

∴ ∠𝐵 = ∠𝐴𝐶𝐵 = 12(180° − ∠𝐴) = 72°

∵ CD bisects ∠𝐴𝐶𝐵

∴ ∠𝐷𝐶𝐵 =2

1∠𝐴𝐶𝐵 = 36°

In △DBC,

∠𝐵𝐷𝐶 = 180 ∘ −∠𝐵 − ∠𝐷𝐶𝐵 = 72 ∘= ∠𝐵

∴ 𝐶𝐷 = 𝐶𝐵

△ 𝐵𝐶𝐷 is an isosceles triangle.

3.5.3 Equilateral Triangles

A. Properties and Criteria of Equilateral Triangles

1．C 2．B 3．C 4．C 5．D 6．60°

7．10cm 8. 90° 9．60° 10．6cm．

11. (1) Proof. ∵ △ABC is an equilateral triangle,

∴ ∠𝐵𝐴𝐸 = ∠𝐶 = 60°, 𝐴𝐵 = 𝐶𝐴.

In △ABE and △CAD,

∵ 𝐴𝐵 = 𝐶𝐴, ∠𝐵𝐴𝐸 = ∠𝐶, 𝐴𝐸 = 𝐶𝐷,

∴ △ 𝐴𝐵𝐸 ≌△ 𝐶𝐴𝐷(𝑆𝐴𝑆).

(2) Solution: ∵ ∠𝐵𝐹𝐷 = ∠𝐴𝐵𝐸 + ∠𝐵𝐴𝐷,

Also, ∵ △ 𝐴𝐵𝐸 ≌△ 𝐶𝐴𝐷,

Page 21 of 37

∴ ∠𝐴𝐵𝐸 = ∠𝐶𝐴𝐷.

∴ ∠𝐵𝐹𝐷 = ∠𝐶𝐴𝐷 + ∠𝐵𝐴𝐷 = ∠𝐵𝐴𝐶 = 60°.

12．Solution: △ 𝐶𝐸𝐵 is an equilateral triangle.

Proof. ∵ 𝐴𝐵 = 𝐵𝐶, ∠𝐴𝐵𝐶 = 120°, 𝐵𝐸 ⊥ 𝐴𝐶,

∴ ∠𝐶𝐵𝐸 = ∠𝐴𝐵𝐸 = 60°.

Also, 𝐷𝐸 = 𝐷𝐵, 𝐵𝐸 ⊥ 𝐴𝐶,

∴ 𝐶𝐵 = 𝐶𝐸.

∴ △ 𝐶𝐸𝐵 is an equilateral triangle.

B. Properties and Applications of Right Triangles with a 𝟑𝟎° Angle

1. D 2. A 3. B 4. 2 5. 2 6. 5 7. 36 3

8．Solution: As shown in the graph, connect DB.

∵ MN perpendicularly bisects AB.

∴ 𝐴𝐷 = 𝐷𝐵, and ∠𝐴 = ∠𝐴𝐵𝐷.

∵ 𝐵𝐴 = 𝐵𝐶, ∠𝐵 = 120 ∘.

∴ ∠𝐴 = ∠𝐶 =2

1(180° − 120°) = 30°.

∴ ∠𝐴𝐵𝐷 = 30°.

Also, ∵ ∠𝐴𝐵𝐶 = 120 ∘,

∴ ∠𝐷𝐵𝐶 = 120° − 30° = 90°

∴ 𝐵𝐷 =2

1𝐷𝐶

∴ 𝐴𝐷 =2

1𝐷𝐶.

Page 22 of 37

9．Solution:

∵ In △ 𝐴𝐵𝐶, ∠𝐶 = 90°，∠𝐴𝐵𝐶 = 60°, 𝐵𝐷 bisects ∠𝐴𝐵𝐶,

∴ ∠2 = ∠3 = 30°.

In 𝑅𝑡 △ 𝐵𝐶𝐷,

𝐶𝐷 =2

1𝐵𝐷，∠4 = 90° − 30° = 60°

∴ ∠1 + ∠2 = 60°

∴ ∠1 = ∠2 = 30°

∴ 𝐴𝐷 = 𝐵𝐷.

∴ 𝐴𝐶 = 𝐴𝐷 + 𝐶𝐷 =2

3𝐴𝐷.

Also, ∵ 𝐴𝐷 = 6

∴ 𝐴𝐶 = 9.

Page 23 of 37

Quiz Yourself (A)


1. C 2. C 3. D 4. D 5. B


6. 4 7. 135° 8. 60° 9. 55° 10. 𝑎 = −2，𝑏 = −3．

11.（1）8cm and 2cm, or 5cm and 5cm

（2）70° and 70°, or 40° and 100°

12.55° 13．80°


14. Proof. ∵ 𝐴𝐶 ∥ 𝐷𝐹,

∴ ∠𝐴 = ∠𝐸𝐷𝐹，

∵ 𝐵𝐶 ∥ 𝐸𝐹，

∴ ∠𝐴𝐵𝐶 = ∠𝐸，

In △ABC and △DEF,

∵ ∠𝐴 = ∠𝐸𝐷𝐹, 𝐴𝐵 = 𝐷𝐸, ∠𝐴𝐵𝐶 = ∠𝐸

∴ △ 𝐴𝐵𝐶 ≅△ 𝐷𝐸𝐹(𝐴𝑆𝐴).

15. Proof. ∵ 𝐴𝐵 = 𝐴𝐶，∴ ∠𝐵 = ∠𝐶．

Also, ∵ ∠𝐴𝐷𝐶 = ∠𝐵 + ∠𝐵𝐴𝐷，∠𝐴𝐷𝐶 = ∠𝐴𝐷𝐸 + ∠𝐶𝐷𝐸,

∴ ∠𝐵 + ∠𝐵𝐴𝐷 = ∠𝐴𝐷𝐸 + ∠𝐶𝐷𝐸.

∵ ∠𝐴𝐷𝐸 = ∠𝐵,

∴ ∠𝐵𝐴𝐷 = ∠𝐶𝐷𝐸,

In △ 𝐴𝐵𝐷 and △ 𝐷𝐶𝐸,

Page 24 of 37

∵ ∠𝐵𝐴𝐷 = ∠𝐶𝐷𝐸，𝐴𝐷 = 𝐷𝐸𝐴𝐷 = 𝐷𝐸,

∴ △ 𝐴𝐵𝐷 ≅△ 𝐷𝐶𝐸(𝐴𝐴𝑆).

16．Solution:

As shown in the graph, passing D construct 𝐷𝐹 ⊥ 𝐵𝐶 at F.

∵ BD bisects ∠𝐴𝐵𝐶, 𝐷𝐸 ⊥ 𝐴𝐵

∴ 𝐷𝐸 = 𝐷𝐹, and

∴ 𝑆 △ 𝐴𝐵𝐶 =1

2𝐴𝐵 ⋅ 𝐷𝐸 +

1

2𝐵𝐶 ⋅ 𝐷𝐹 = 90.

That is, 1

2× 18 ⋅ 𝐷𝐸 +

1

2× 12 ⋅ 𝐷𝐸 = 90.

Thus, 𝐷𝐸 = 6.

17. Solution: 𝑃𝐸 = 𝑃𝐹.

Reasoning: Passing P construct 𝑃𝑀 ⊥ 𝑂𝐴 at M and 𝑃𝑃𝑁 ⊥ 𝑂𝐵 at N.

Then ∠𝑃𝑀𝐸 = ∠𝑃𝑁𝐹 = 90°.

∵ OP bisects ∠𝐴𝑂𝐵,

∴ 𝑃𝑀 = 𝑃𝑁.

∵ ∠𝐴𝑂𝐵 = ∠𝑃𝑀𝐸 = ∠𝑃𝑁𝐹 = 90°.

∴ ∠𝑀𝑃𝑁 = 90°.

∵ ∠𝐸𝑃𝐹 = 90°,

∴ ∠𝑀𝑃𝐸 = ∠𝐹𝑃𝑁.

In △ 𝑃𝐸𝑀 and △ 𝑃𝐹𝑁,

∵ ∠𝑃𝑀𝐸 = ∠𝑃𝑁𝐹，𝑃𝑀 = 𝑃𝑁，∠𝑀𝑃𝐸 = ∠𝑁𝑃𝐹

∴ △ 𝑃𝐸𝑀 ≌△ 𝑃𝐹𝑁.

∴ 𝑃𝐸 = 𝑃𝐹.

Page 25 of 37

18. Solution.

(1) ∵ △ABC is an equilateral triangle,

∴ ∠𝐵 = 60°.

∵ 𝐷𝐸 ∥ 𝐴𝐵,

∴ ∠𝐸𝐷𝐶 = ∠𝐵 = 60°.

∵ 𝐸𝐹 ⊥ 𝐷𝐸,

∴ ∠𝐷𝐸𝐹 = 90°.

∴ ∠𝐹 = 90° − ∠𝐸𝐷𝐶 = 30°.

(2) ∵ ∠𝐴𝐶𝐵 = 60°，∠𝐸𝐷𝐶 = 60°，

∴ △EDC is an equilateral triangle.

∴ 𝐸𝐷 = 𝐷𝐶 = 2.

∵ ∠𝐷𝐸𝐹 = 90°, ∠𝐹 = 30 ∘,

∴ 𝐷𝐹 = 2𝐷𝐸 = 4.

Quiz Yourself (B)

1. C 2. A 3. C 4. D 5. B 6. 4 7. 60°

8．

9. (1) Proof. ∵ △ 𝐴𝐵𝐶 is an equilateral triangle.

∴ 𝐴𝐵 = 𝐶𝐴 = 𝐵𝐶, ∠𝐵𝐴𝐸 = ∠𝐴𝐶𝐷 = 60°.

In △ 𝐴𝐵𝐸 and △ 𝐶𝐴𝐷,

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∵ 𝐴𝐵 = 𝐶𝐴, ∠𝐵𝐴𝐸 = ∠𝐴𝐶𝐷 = 60°, 𝐴𝐸 = 𝐶𝐷，

∴ △ 𝐴𝐵𝐸 ≌△ 𝐶𝐴𝐷(𝑆𝐴𝑆),

∴ 𝐴𝐷 = 𝐵𝐸.

(2) Solution: ∵ △ 𝐴𝐵𝐸 ≅△ 𝐶𝐴𝐷,

∴ ∠𝐶𝐴𝐷 = ∠𝐴𝐵𝐸.

∴ ∠𝐵𝑃𝑄 = ∠𝐴𝐵𝐸 + ∠𝐵𝐴𝐷 = ∠𝐵𝐴𝐷 + ∠𝐶𝐴𝐷 = ∠𝐵𝐴𝐸 = 60°.

∵ 𝐵𝑄 ⊥ 𝐴𝐷,

∴ ∠𝐴𝑄𝐵 = 90°.

∴ ∠𝑃𝐵𝑄 = 90° − 60° = 30°.

∵ 𝑃𝑄 = 3,

∴ In 𝑅𝑡 △ 𝐵𝑃𝑄, 𝐵𝑃 = 2𝑃𝑄 = 6.

Also, ∵ 𝑃𝐸 = 1,

∴ 𝐴𝐷 = 𝐵𝐸 = 𝐵𝑃 + 𝑃𝐸 = 6 + 1 = 7．

10. Proof. ∵ △ 𝐴𝐵𝐷 and △ 𝐴𝐶𝐸 are both equilateral triangles.

∴ 𝐴𝐶 = 𝐴𝐸，𝐴𝐷 = 𝐴𝐵.

∵ ∠𝐸𝐴𝐶 = ∠𝐷𝐴𝐵 = 60°, and ∠𝐸𝐴𝐶 + ∠𝐵𝐴𝐶 = ∠𝐷𝐴𝐵 + ∠𝐵𝐴𝐶.

That is, ∠𝐸𝐴𝐵 = ∠𝐶𝐴𝐷．

In △ 𝐸𝐴𝐵 and △ 𝐶𝐴𝐷,

𝐴𝐸 = 𝐴𝐶，∠𝐸𝐴𝐵 = ∠𝐶𝐴𝐷，𝐴𝐵 = 𝐴𝐷,

∴ △ 𝐸𝐴𝐵 ≌△ 𝐶𝐴𝐷.

∴ 𝐵𝐸 = 𝐶𝐷.

Page 27 of 37

11. Proof. (1) From the given information, we know

∠𝐴 + ∠𝐵 = 90°𝑎𝑛𝑑 ∠𝐴 = ∠𝐷.

∴ ∠𝐷 + ∠𝐵 = 90°.

∴ 𝐴𝐵 ⊥ 𝐷𝐸.

(2) ∵ 𝐴𝐵 ⊥ 𝐷𝐸, 𝐴𝐶 ⊥ 𝐵𝐷

∴ ∠𝐵𝑃𝐷 = ∠𝐴𝐶𝐵 = 90°.

In △ 𝐴𝐵𝐶 and △ 𝐷𝐵𝑃,

∵ ∠𝐴 = ∠𝐷∠𝐴𝐶𝐵 = ∠𝐷𝑃𝐵𝐵𝐶 = 𝐵𝑃,

∴ △ 𝐴𝐵𝐶 ≌△ 𝐷𝐵𝑃(𝐴𝐴𝑆).

Page 28 of 37

Chapter 4 Pythagorean Theorem

4.1 Pythagorean Theorem

A. Pythagorean Theorem and Its Representation

1. 15, 20

2.（1） 21 （2）2 21，8 21 （3） 22

3. 12𝑐𝑚 𝑜𝑟 7 + 7 𝑐𝑚

4. （1）25𝑐𝑚²，20𝑐𝑚 (2) 72𝜋𝑐𝑚²，24 + 12𝜋𝑐𝑚

5. 2.2 m

B. Application to Pythagorean Theorem

1. D 2. D 3．2 + 22 𝑐𝑚 4. 24 5. 10𝑘𝑚

C. Representing Irrational Numbers on Number Line Using

Pythagorean Theorem

1.（1）

（2）

（3）

2. Answers are shown on the graph below.

Page 29 of 37

(1) (4) (2) (3)

4.2 Converse Theorem of the Pythagorean Theorem

1. A 2. A 3．Right triangle, ∠𝐴 = 30°

4. 2.4 5. 3 6. Right triangle

7. Isosceles triangle or right triangle 8. 13 9. Right triangle

4.3 Applications to Pythagorean Theorem and Its Converse

A. Applications to Pythagorean Triple

1. C 2. B 3. (1）17 （2）9 （3）24 （4）25．

4. 15𝑘𝑚. 5. 100𝑐𝑚. 6. 20𝑚 7. 6𝑚

B. Applications to Special Triangles

1. The area is 2

31+. The perimeter is 323 ++ .

2. The length is 30． 3. The area of the quadrilateral is 3924+ .

4. Extend AD to point E such that 𝐴𝐷 = 𝐷𝐸. Connect BE.

Prove that △ 𝐴𝐷𝐶 ≌△ 𝐸𝐷𝐵 to obtain 𝐵𝐸 = 𝐴𝐶 = 32 .

Using Pythagorean Theorem to prove that △ 𝐴𝐵𝐸 is a right triangle, and

Page 30 of 37

∠𝐵𝐸𝐴 = 90° to obtain 𝐵𝐷 =2

1𝐴𝐸 = 3, 𝐵𝐶 = 6 . The area of △ 𝐴𝐵𝐶

and the area of △ 𝐴𝐵𝐸 are both 12.

5. (1) The length of CE is 3. (2) The area of △ 𝐴𝐸𝐷 is 6.

C. Applications of Rectangular Coordinate System

1. 5 2. (3, 4), (2, 4), (8, 4) 3. The distance is 2√5.

4.4 Rotation Transformation

A. Definition of Rotation and Its Properties

1. A 2. C 3. C 4. 60°

5. Rotate △ 𝐵𝐶𝐸 about point B counterclockwise 60° and it will coincide

with △ 𝐵𝐴𝐹.

B. Reasoning and Calculation on Rotating 𝟗𝟎° on a Triangle About a

Point

1. 90° 2. √2 3. (−3, −2)

4．(1) The complete graph is shown below.

Page 31 of 37

(2) Using the properties of rotation we obtain ∠𝐷𝐶𝐹 = 90 ∘.

∴ ∠𝐷𝐶𝐸 + ∠𝐸𝐶𝐹 = 90°.

∵ ∠𝐴𝐶𝐵 = 90°,

∴ ∠𝐷𝐶𝐸 + ∠𝐵𝐶𝐷 = 90°.

∴ ∠𝐸𝐶𝐹 = ∠𝐵𝐶𝐷.

∵ 𝐸𝐹 ∥ 𝐷𝐶,

∴ ∠𝐸𝐹𝐶 + ∠𝐷𝐶𝐹 = 180°.

∴ ∠𝐸𝐹𝐶 = 90°.

In △ 𝐵𝐷𝐶 and △ 𝐸𝐹𝐶,

∵ 𝐷𝐶 = 𝐹𝐶, ∠𝐵𝐶𝐷 = ∠𝐸𝐶𝐹, 𝐵𝐶 = 𝐸𝐶,

∴ △ 𝐵𝐷𝐶 ≅△ 𝐸𝐹𝐶(𝑆𝐴𝑆).

∴ ∠𝐵𝐷𝐶 = ∠𝐸𝐹𝐶 = 90°.

Page 32 of 37

Quiz Yourself (A)


1. B 2. C 3. A 4. C 5. B

6. C 7. B 8. A 9. B 10. B


11. 120 degrees 12. 10𝑐𝑚 13. 336 14. 100

15. 7 16. 5√2 17. 3400bahts


18．（1）

（2）6，135°.

19．Proof. ∵ △ABC is an isosceles triangle.

∴ 𝐴𝐵 = 𝐵𝐶, ∠𝐴 = ∠𝐶.

∵ Rotate △ 𝐴𝐵𝐶 about point B counterclockwise for α degrees to the

location of △ 𝐴1𝐵1𝐶1.

∴ 𝐴1𝐵 = 𝐴𝐵 = 𝐵𝐶，∠𝐴 = ∠𝐴1 = ∠𝐶，∠𝐴1𝐵𝐷 = ∠𝐶𝐵𝐶1.

In △ 𝐵𝐶𝐹 and △ 𝐵𝐴1𝐷,

∠𝐴1 = ∠𝐶，𝐴1𝐵 = 𝐵𝐶，∠𝐴1𝐵𝐷 = ∠𝐶𝐵𝐹.

∴ △ 𝐵𝐶𝐹 ≌△ 𝐵𝐴1𝐷.

20．Solution：Label the points as shown in the graph.

Let 𝐴𝐶 = 𝑥. Then 𝐵𝐶 = (70 − 𝑥)𝑐𝑚 when the rope is pulled straight.

Page 33 of 37

If angle C is 90 degrees, by Pythagorean Theorem, we have

502 = 𝑥2 + (70 − 𝑥)2.

Solving the equation we obtain 𝑥 = 40𝑐𝑚 𝑜𝑟 30𝑐𝑚.

If AC is the hypotenuse, then

502 + (70 − 𝑥)2 = 𝑥2,

Solving the equation we obtain 𝑥 = 526

7𝑐𝑚.

If BC is the hypotenuse, then

502 + 𝑥2 = (70 − 𝑥)2.

Solving the equation we obtain 𝑥 = 171

7𝑐𝑚.

21．15𝑚 22. 216𝑚²

Quiz Yourself (B)

1. B 2. C 3. C 4. C 5. A

6. C (Note: It can be an acute or an obtuse triangle.)

7. 2√3𝑐𝑚 8. 1.5 9. 4 10. 49𝑐𝑚2

11. 0.8𝑚 12. 𝐴𝐵 = 𝐴𝐶 = 5, 𝐵𝐶 = 6

13. ∵ 𝐴𝐵 = 100𝑘𝑚, 𝐴𝐷 = 60𝑘𝑚

∴ In 𝑅𝑡 △ 𝐴𝐵𝐷, by Pythagorean Theorem,

𝐵𝐷 = √𝐴𝐵2 − 𝐴𝐷2 = 80𝑘𝑚.

The typhoon will move from point B to point D in 80 ÷ 20 = 4 hours.

∵ Assume that everything within 30 km of the center of the typhoon

is in danger.

Page 34 of 37

∴ People should be evacuated before the typhoon reaches Location E as

shown in the graph.

∵ 𝐵𝐸 = 𝐵𝐷 − 𝐷𝐸 = 80 − 30 = 50𝑘𝑚.

∴ To avoid the danger, the latest time for people at the location D to

evacuate is within 50 ÷ 20 = 2.5 hours.

14. Solution:

(1) In △ABC, since 𝐴𝐶 = 𝑀𝐴 = 1, and 𝐴𝐵 = 𝑥, then 𝐵𝐶 = 3 − 𝑥.

We can set up a system of equations that 𝑥 must satisfy as the following.

{1 + 𝑥 > 3 − 𝑥1 + 3 − 𝑥 > 𝑥.

Solving this system we can obtain the range of 𝑥: 1 < 𝑥 < 2.

(2) ① If AC is the hypotenuse, then 1 = 𝑥2 + (3 − 𝑥)2.

That is, 𝑥2 − 3𝑥 + 4 = 0. There is no solution for this equation.

② If AB is the hypotenuse, then 𝑥2 = (3 − 𝑥)2 + 1. Solving this system

we can obtain 𝑥 =5

3, satisfying 1 < 𝑥 < 2.

③ If BC is the hypotenuse, then (3 − 𝑥)2 = 1 + 𝑥2. Solving this system

we can obtain 𝑥 =4

3, satisfying 1 < 𝑥 < 2.

Thus, the solutions are 𝑥 =5

3 or 𝑥 =

4

3.

Page 35 of 37

Chapter 5 Statistics-Organizing and Handling Data

5.1 Arithmetic Mean and Weighted Mean

5.1.1 Meaning of Arithmetic Mean

1. 1

2 2. 9.7 3. 8，2

4.（1）−1 （2）5 5. 1300bahts

5.1.2 Weighted Mean

1. C 2. 900 3. 53 4. 82.4 5. 4 6. Student A

5.1.3 Construction and Interpretation of Pie Charts

1.（1）Candidate A （2）Candidate B

2. Size M sells the most, they should be stocked the most. Size XXL sells

the least, they should be stocked the least.

5.2 Applications to Mean, Median, and Mode

5.2.1 Median and Mode

1. 3.5, 1, 5 2. 5 or 9 3. 30, 42 4. C

5.（1） approximately 9958 （2）Inappropriate

5.2.2 Proper Choice from Mean, Median, and Mode

1. B 2. A 3. 6，6，6

4.（1）147 （2）It is less than the median: 147. So it is not so ideal.

Page 36 of 37

5. Mean：20 Mode：21 Median：21

5.3 Range, Variance, and Standard Deviation

5.3.1 Measure of Dispersion for Discrete Data

1. D 2. C 3. B 4. 4, 3 5. Answer not provided.

5.3.2 Decision Making Based on Variance

1.（1）Class 2 （2）Class 1

2.（1）Package A has a mean：504.8, and a variance：15.76. Package B

has a mean：504.8, and a variance：5.56

（2）B

Page 37 of 37

Quiz Yourself (A)


1. D 2. D 3. B 4. A 5. C 6. B 7. D 8. C


9. 81 10. B 11. 6 12. 2


13. The mean annual profit of each employee is 8 million baht.

14. The 1st prize: Participant B. The 2nd prize: Participant A.

15.（1）Sample size：50，𝑚 = 5，median：28 （2）300 students

16.（1）8，8，9，8 （2）A, because A’s variance is smaller.

Quiz Yourself (B)

1. B 2. D 3. C 4. D 5. D 6. 39，40

7. 1，3，5 or 2，3，4 8. 2 9. 34

10.（1）The mean score for A is 9, and for B is 9.

（2）The variance for A is 2

3, and for B is

4

3.

（3）A should be selected because its variance is smaller and so is more

stable.

11.（1）Mean：320 Median：210 Mode：210

（2）No, it is inappropriate.

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