1

Exercise Solutions

Exercise 1

a) i. Reflection

ii. Rotation

iii. Rotation

b)

x

y

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

2

6

4

-2

-4

-6

-2-4-6 2 4 6

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

c)

x

y

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

2

6

4

-2

-4

-6

-2-4-6 2 4 6

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

Exercise 2

a) i. Translation

ii. Neither

iii. Resizing

2

b)

x

y

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

2

6

4

-2

-4

-6

-2-4-6 2 4 6

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

c)

x

y

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

2

6

4

-2

-4

-6

-2-4-6 2 4 6

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

Exercise 3

i)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

ii)

x

y

2

4

-2

-4

-2-4 2 4

iii)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

iv)

x

y

2

4

-2

-4

-2-4 2 4

3

Exercise 4

i) Resize by a factor of 2.Translate (x + 4, y + 2).

ii) Resize by a factor of 12.

Clockwise rotation of 90◦ about the origin.Translate (x− 2, y − 2).

Problem Set Solutions

1.

line 1

line 2

2. The actual time is 5:10.

3. (a)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

(b)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

(c)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

(d)

x

y

2

6

4

-2

-4

-6

-2-4-6 2 4 6

4. Reflect with mirror line x = −2.Clockwise rotation of 90◦ about origin.

4

5. (a)

x

y

1

3

2

-1

-2

-3

-1-2-3 1 2 3

4

5

-5

-4

4 5-5 -4

(b)

x

y

1

3

2

-1

-2

-3

-1-2-3 1 2 3

4

5

-5

-4

4 5-5 -4

(c)

x

y

1

3

2

-1

-2

-3

-1-2-3 1 2 3

4

5

-5

-4

4 5-5 -4

(d)

x

y

1

3

2

-1

-2

-3

-1-2-3 1 2 3

4

5

-5

-4

4 5-5 -4

6. Counterclockwise rotation of 90◦ about the origin.Translate (x + 6, y + 6).

7. Translate: (x + 10, y + 30)(x + 20, y − 20)(x + 20, y)(x + 10, y + 50)(x + 20, y − 10)(x + 20, y − 30)

8. The largest product is 6 × 9 × 11 = 594

9. The answer is (d)

10. Since GHFD is a square, we know that HF and GH must be the same length. In otherwords, the length must be a common factor of 21 and 28. The only common factorsof 21 and 28 are 1 and 7. This means the area of GHFD is either 1 × 1 = 1 cm2 or7 × 7 = 49 cm2. Since the area of GHFD is greater than the area of the other rectanglesthe area of GHFD must be 49 cm2.

5

11. The inside faces that are touching a face of another die must be 3 or higher because weneed them to add to 9. Using this rule we know the arrangements of the inside faces. Forthe die on the corner, we know where the 4 and 5 are, so the 6 must go on the green face.Therefore, the face with the “p” is 3.

12. Filling in the grid, we have

20

23 14

456

7

21

1 3

9 8 2

222425

16

17

15

12

10

11

13

19

18

Therefore the missing number is 14.

6

13. On the first fold, we can see that AB has the same length as CP , which means the lengthof PQ is 8 − 5 = 3 cm. We have the following, which we can then halve.

3.000

2.000

3.000

5.000

Would halve into

1.0004.000

1.000

3.000

M L

D

R

Q

C

7

The area of DRQC is the area of DMLC minus the areas of the two triangles DMR andRLQ.

AreaDRQC = AreaDMLC − AreaDMR − AreaRLQ

= (5)(4) − (4)(4)

2− (1)(1)

2= 20 − 8 − 0.5

= 11.5

∴ The area of DRQC is 11.5cm2.