DiiponeAcademy
CAPS · MATHEMATICS · GRADE 8

PRACTICE WORKSHEET

SET #482913TOTAL: 180 MARKS
NAME:DATE:
  1. Answer ALL the questions.
  2. Round answers to TWO decimal places, unless stated otherwise.
  3. Where a question has more than one possible answer, give ALL possible answers.
  4. Provide reasons for your statements in geometry questions.
  5. Diagrams are not necessarily drawn to scale.
  6. You may use an approved scientific calculator, unless the question states otherwise.
1.
Gr 8 · Properties of operations · K

Name the property of operations illustrated by: p×(q+r)=p×q+p×rp\times(q+r)=p\times{}q+p\times{}r

(1)
2.
Gr 8 · Real number system (ℕ, ℕ₀, ℤ, ℚ, ℚ′) · K

Which number sets does 81\sqrt{81} belong to? Consider natural numbers (ℕ), whole numbers (ℕ0_{0}), integers (ℤ), rational numbers (ℚ) and irrational numbers (ℚ′).

(2)
3.
Gr 8 · Factors, HCF & LCM · R

Determine the HCF of 72 and 64 by using prime factors.

(3)
4.
Gr 8 · Exponents & roots · R

Simplify, leaving the answer as a power of 5: 54×535^{4}\times5^{3}

(2)
5.
Gr 8 · Scientific notation (large numbers) · K

Write as an ordinary number: 6,3×1046{,}3\times10^{4}

(2)
6.
Gr 8 · Common fractions · R

Write down the reciprocal of 3. Check your answer by multiplying the two together.

(2)
7.
Gr 8 · Decimal fractions · R

Fill in < or > to make the statement true: 0,459 □ 0,022

(1)
8.
Gr 8 · Percentages · R

A spaza shop sold 1160 cold drinks last week. This week sales decrease by 10%. How many cold drinks are sold this week?

(3)
9.
Gr 8 · Integers · K

Calculate EACH of the following: (a)(10)2(a)(-10)^{2} (b)102(b)-10^{2}

(2)
10.
Gr 8 · Like terms · R

Simplify: 24x5÷4x324x^{5}\div4x^{3}

(2)
11.
Gr 8 · Equations · R

Solve for x: x/3=83=-8

(1)
12.
Gr 8 · Number patterns · R

A row of octagons is built from matchsticks so that each new octagon shares one side with the last. n: 1; 2; 3; 4 Matchsticks: 8; 15; 22; ?

12.1

Complete the table by writing down the number of matchsticks needed for n=4n=4 octagons.

(2)
12.2

Describe the rule of the pattern in words.

(1)
12.3

Write down the general term Tₙ (the number of matchsticks for n octagons).

(2)
12.4

How many matchsticks are needed for a row of 15 octagons?

(2)
[7]
13.
Gr 8 · Functions & relationships · R

The rule y=8x2y=8x-2 connects x and y in the table below. Determine the TWO missing values. x: 1; 2; 3; 5; ? y: ?; 14; 22; 38; 134

x× 8− 2y
(3)
14.
Gr 8 · Expand & simplify · R

Simplify: (20x5+36x316x2)÷4x2(20x^{5}+36x^{3}-16x^{2})\div4x^{2}

(3)
15.
Gr 8 · Equations from words · C

The length of a rectangle is 6 cm longer than its breadth. The perimeter of the rectangle is 68 cm. Determine the length and the breadth of the rectangle.

x + 6x
(4)
16.
Gr 8 · Interpret graphs · C

Naledi cycles from home to the market and then on to the train station. The distance–time graph shows the whole journey.

0102030405060700481216Distance from home (km)Time (minutes)
16.1

How far from home was Naledi 40 minutes after leaving?

(1)
16.2

How long did Naledi stay at the market?

(1)
16.3

Calculate Naledi's average speed, in km/h, on the ride from home to the market.

(3)
[5]
17.
Gr 8 · Global graph features · K

The graph shows the water level plotted against minutes. Read off the graph: what is the water level at 1 on the horizontal axis?

012345678020406080100Water in the tank (ℓ)Time (minutes)
(2)
18.
Gr 8 · Substitution · R

If x=36x=36, evaluate: 9x\sqrt{9x}

(2)
19.
Gr 8 · Financial maths · R

Pieter invests R1 900 at 6% simple interest per year for 6 years. How much money will be in the account at the end of the 6 years?

(3)
20.
Gr 8 · Ratio & proportion · C

Rangers catch and tag 105 impala in a private game reserve, then release them. Later they catch a second sample of 57 impala and find that 15 of them are tagged. Estimate the total impala population.

(3)
21.
Gr 8 · Problem solving · P

The length of a rectangle is 4 cm more than its breadth. The perimeter of the rectangle is 28 cm. Calculate the AREA of the rectangle.

(5)
22.
Gr 8 · Data handling · R

The Tzaneen netball team scored the following numbers of goals in its last 8 matches: 25; 33; 31; 28; 33; 45; 29; 24.

22.1

Calculate the mean number of goals scored per match.

(2)
22.2

Determine the median of the data.

(2)
22.3

Write down the mode of the data, and give a reason for your answer.

(1)
22.4

Determine the range of the data.

(1)
22.5

The value 45 is an outlier. If it is removed from the data set, which changes more: the mean or the median? Show calculations for BOTH.

(4)
[10]
23.
Gr 8 · Interpret charts · R

The broken-line graph shows the height (in cm) of a bean plant measured at the end of each week.

0246810121416Height (cm)Week 1Week 2Week 3Week 4Week 5Week 6
23.1

How tall was the bean plant at the end of Week 5?

(1)
23.2

Between which two consecutive points did the height increase the most? What was that increase?

(2)
23.3

Describe the overall trend from Week 1 to Week 6.

(2)
[5]
24.
Gr 8 · Sampling · K

Two methods are suggested to test 21 tins from a production batch of 3453 tins for a factory quality check. Method A: The quality inspector only tests the 21 tins packed last, at the end of the shift. Method B: A computer randomly selects 21 tin numbers from the whole batch. Which method is RANDOM sampling? Give a reason.

(2)
25.
Gr 8 · Venn diagrams · C

Of the 34 learners in Grade 8D, 19 are in set Cricket and 20 are in set Hockey; 8 learners are in both sets. How many learners are in NEITHER set?

8DCricket (19)Hockey (20)??8?
(3)
26.
Gr 8 · Probability · K

A spinner is divided into 4 equal sectors, each a different colour: yellow; purple; blue; green. The spinner is spun once. Answer the questions below about the sample space.

yellowpurplebluegreen
26.1

Write down the sample space S (the set of all possible outcomes).

(2)
26.2

State n(S)n(S), the number of outcomes in the sample space.

(1)
26.3

Each outcome is equally likely. Write down the probability of one outcome, then show that the probabilities of all the outcomes add up to 1.

(2)
[5]
27.
Gr 8 · Financial & rate chains · C

The Mokoena family drive to East London at a constant average speed. The average speed is 60 km/h. The trip takes 150 minutes. The car uses 8 ℓ per 100 km. How many litres of petrol does the trip use?

(4)
28.
Gr 8 · Finance & rates revision (multi-step) · C

Mr Sithole drives to Kimberley at a constant average speed. The average speed is 80 km/h. The trip takes 180 minutes. The car uses 10 ℓ per 100 km. Petrol costs R21,95 per litre.

28.1

How long does the trip take, in hours?

(2)
28.2

Hence, how far is the trip?

(2)
28.3

Hence calculate the petrol cost for the trip.

(3)
[7]
29.
Gr 8 · Numeric evaluation chains · C

Calculate, without using a calculator: 18(911)×(16)÷(27311)18-(-\tfrac{9}{11})\times(-16)\div(27-\tfrac{3}{11})

(4)
30.
Gr 8 · Data reasoning chains · C

Lwandle recorded the number of minutes spent on homework on each of 7 evenings: 24; 10; 35; 19; 32; 37; 33. On the next evening, Lwandle spent 14 minutes on homework. In total, there are now 8 evenings. Calculate the mean number of minutes spent on homework per evening for all 8 evenings.

(3)
31.
Gr 8 · Geometry reasoning chains · C

△RST is similar to △XYZ (the triangles are equiangular). RS=4RS=4 cm, ST=18ST=18 cm and XY=6XY=6 cm. Calculate the length of YZ.

RST4 cm18 cmXYZ6 cm?
(3)
32.
Gr 8 · Transformation sequences · C

On the grid, the line segment AB has endpoints A(4; −6) and B(6; −6), as shown. The segment is first reflected in the line y=xy=x; a second transformation then maps the result onto the dashed image A″B″, with A landing on A″(6; −4). Describe the second transformation fully. (It is a single reflection, rotation or enlargement.)

xy−6−6−4−4−2−222446ABA″B″
(3)
33.
Gr 8 · Algebra chains (expand, solve, translate) · C

Solve for x: 5x8=3x25x-8=3x-2

(3)
34.
Gr 8 · Spot & correct the error · C

Thabo says the hypotenuse of a right triangle with legs 6 cm and 4 cm is 10 cm. Identify the mistake.

(2)
35.
Gr 8 · Always, sometimes or never true? · C

Is this statement ALWAYS, SOMETIMES or NEVER true? Give a reason. "The solution of x+14=11x+14=11 is a positive number."

(2)
36.
Gr 8 · Explain & discuss (teacher-marked) · C

Explain why 1m21m^{2} is NOT equal to 100 cm2^{2}, and give the correct number of cm2^{2} in 1m21m^{2}.

(2)
37.
Gr 8 · Multi-step: real life · C

The Naidoo family drive to East London, keeping an average speed of 100 km/h for 1 h 30 min.

37.1

How far is the trip?

(2)
37.2

The car uses 6 ℓ of petrol per 100 km. How many litres does the trip use?

(2)
37.3

Petrol costs R21,95 per litre. Calculate the petrol cost for the trip.

(2)
[6]
38.
Gr 8 · Angles & parallel lines · C

In the diagram, AB ∥ CD and a transversal cuts both lines. Two angles are marked 5x+275x+27^\circ and 3x+453x+45^\circ.

ABCD5x + 27°3x + 45°
38.1

Calculate the value of x, giving a reason for your equation.

(3)
38.2

Hence calculate the size of each marked angle.

(1)
[4]
39.
Gr 8 · Perimeter & area · R

The figure shows a semicircle with radius 14 cm.

14 cm
39.1

Calculate the perimeter of the semicircle, correct to two decimal places.

(3)
39.2

Calculate the area of the semicircle, correct to two decimal places.

(2)
[5]
40.
Gr 8 · Triangle geometry · R

In △XYZ, side ZX is extended beyond X, forming an exterior angle of 118°. Y^=76\hat{Y}=76^\circ. Calculate, with a reason, the size of x (the angle at Z).

ZXYx76°118°
(2)
41.
Gr 8 · Quadrilaterals · R

The diagram shows quadrilateral PQRS with its equal sides, parallel sides and right angles marked. Name the quadrilateral.

PQRS
(2)
42.
Gr 8 · Congruence & similarity · C

The diagram shows △DEF and △PQR with matching marks. Which condition (SSS, SAS, AAS or RHS) proves that △DEF ≡ △PQR? List the matching pairs.

DEFPQR
(2)
43.
Gr 8 · Constructions · K

In a construction, EG bisects ∠DEF=120DEF=120^\circ. Write down the size of ∠GEF.

(1)
44.
Gr 8 · 3D objects · R

A decagonal pyramid has 20 edges and 11 vertices. Use Euler's formula, F+VE=2F+V-E=2, to determine how many faces it has.

(3)
45.
Gr 8 · Transformations · R

On the grid, △ABC has vertices A(−1; −3), B(−1; −1), C(−5; −2), as shown.

xy−5−5−3−3−1−111335ABC
45.1

Write down the coordinates of A′, B′ and C′, the image of △ABC after a translation of 3 units to the right and 2 units up.

(3)
45.2

Write down the general rule in the form (x; y) → (…).

(2)
45.3

Is the image congruent to △ABC? Give a reason.

(1)
[6]
46.
Gr 8 · Polygons · K

An octagon is drawn below. How many sides does it have?

ABCDEFGH
(1)
47.
Gr 8 · Symmetry · K

A regular heptagon is drawn below. How many lines of symmetry does it have?

(1)
48.
Gr 8 · Coordinates on the Cartesian plane · K

Write down the coordinates of P and Q, shown on the grid.

xy−6−6−4−4−2−222446PQ
(2)
49.
Gr 8 · Parts of a circle · K

The radius of a circular swimming pool is 6 m. Calculate its diameter.

6 m
(1)
50.
Gr 8 · Geometry vocabulary (angles, lines, points) · K

At point M, ray ME and ray MG meet to form an angle. Name this angle using three letters, with the vertex in the middle.

EGM
(1)
51.
Gr 8 · Unit conversions · K

Convert: 370 cm to m

(2)
52.
Gr 8 · Surface area & volume · R

Each edge of the cube in the diagram is 3 cm long.

3 cm3 cm3 cm
52.1

Calculate the total surface area of the cube.

(3)
52.2

Calculate the volume of the cube.

(2)
[5]
53.
Gr 8 · Theorem of Pythagoras · R

In △RST, S^=90\hat{S}=90^\circ, RT=10RT=10 cm and RS=8RS=8 cm. Calculate the length of ST.

RST10 cm8 cm
(3)
54.
Gr 8 · Angle chase (parallel lines) · C

In the diagram, ℓ3_{3} ∥ ℓ4_{4}. The transversals ℓ1_{1} and ℓ2_{2} cut the parallel lines. The marked angles are 48° and 111°. Calculate, with reasons, the size of z.

111°48°rnzℓ₃ℓ₄ℓ₁ℓ₂
(4)
55.
Gr 8 · Composite Pythagoras · C

Two right-angled triangles stand back-to-back on a straight base, sharing the altitude drawn perpendicular to the base. The marked hypotenuse measures 25 cm, and the two pieces of the base measure 7 cm and 10 cm. Calculate the length marked v.

nv7 cm10 cm25 cmℓ₄ℓ₃ℓ₂ℓ₁
(3)
56.
Gr 8 · Revision scene · C

In the figure, ℓ2_{2} ∥ ℓ1_{1}, and ℓ3_{3} ∥ ℓ4_{4}. The transversals ℓ5_{5} and ℓ6_{6} cut the parallel lines. The marked angles are 25°, 104°, 104°, 129° and 46°. Calculate the value of each marked unknown, giving a reason for every answer.

104°104°129°25°46°bmrzpℓ₂ℓ₁ℓ₃ℓ₄ℓ₅ℓ₆
56.1

Calculate, with reasons, the size of b.

(2)
56.2

Calculate, with reasons, the size of m.

(2)
56.3

Calculate, with reasons, the size of r.

(2)
56.4

Calculate, with reasons, the size of z.

(2)
56.5

Calculate, with reasons, the size of p.

(2)
[10]
TOTAL: 180 MARKS
MEMORANDUM · SET #482913

MARKING GUIDELINE

1.
∴ The distributive property
(1)
2.
81\sqrt{81} belongs to: ℕ, ℕ0_{0}, ℤ, ℚ
(2)
3.
72=23×3272=2^{3}\times3^{2}
64=2664=2^{6}
HCF=23=8HCF=2^{3}=8
(3)
4.
=54+3=5^{4+3}
=57=5^{7}
(2)
5.
=6,3×10000=6{,}3\times10\,000
∴ = 63 000
(2)
6.
3=313=\tfrac{3}{1}
∴ Reciprocal =13=\tfrac{1}{3}
Check: 31×13=33=1\tfrac{3}{1}\times\tfrac{1}{3}=\tfrac{3}{3}=1
(2)
7.
0,459>0,0220{,}459>0{,}022
(1)
8.
Decrease =10%×1160=116=10\%\times1160=116
∴ New number = 1160 − 116 = 1044 cold drinks
(3)
9.
(a)(10)2=(10)×(10)=100(a)(-10)^{2}=(-10)\times(-10)=100
(b)102=(10×10)=100(b)-10^{2}=-(10\times10)=-100
(2)
10.
=(24÷4)x53=(24\div4)x^{5-3}
=6x2=6x^{2}
(2)
11.
x=8×3x=-8\times3
x=24x=-24
(1)
12.1
Matchsticks(4)=7×4+1(4)=7\times4+1
∴ = 29 matchsticks
(2)
12.2
∴ Each new octagon adds 7 matchsticks to the total, starting at 8 for n=1n=1.
(1)
12.3
Tₙ =8+(n1)×7=8+(n-1)\times7
∴ Tₙ =7n+1=7n+1
(2)
12.4
T=7×15+1T=7\times15+1
T=106T=106 matchsticks
(2)
13.
Missing y (x = 1): y=8(1)2=6y=8(1)-2=6
Missing x: 8x2=1348x-2=134
8x=134+2=1368x=134+2=136
x=17x=17
(3)
14.
=20x5=20x^{5}/4x2+36x34x^{2}+36x^{3}/4x216x24x^{2}-16x^{2}/4x24x^{2}
=5x3+9x4=5x^{3}+9x-4
(3)
15.
Let the breadth =x=x cm, so the length = (x + 6) cm
2(x+x+6)=682(x+x+6)=68
4x+12=684x+12=68
4x=6812=564x=68-12=56
x=14x=14: breadth = 14 cm and length = 14 + 6 = 20 cm
(4)
16.1
∴ 8 km from home
(1)
16.2
∴ Stop = 40 − 20 = 20 minutes
(1)
16.3
Time = 20 minutes =2060=13=\tfrac{20}{60}=\tfrac{1}{3} hour
Average speed = distance ÷\div time =8÷13=8\div\tfrac{1}{3}
∴ Average speed = 24 km/h
(3)
17.
Go to 1 on the horizontal axis, move straight up to the graph, then straight across to the vertical axis
∴ 40 ℓ
(2)
18.
=9×36=324=\sqrt{9\times36}=\sqrt{324}
∴ = 18
(2)
19.
Interest for 1 year =R1900×6100=R114,00=\mathrm{R}1\,900\times\tfrac{6}{100}=\mathrm{R}114{,}00
Interest for 6 years =R114,00×6=R684,00=\mathrm{R}114{,}00\times6=\mathrm{R}684{,}00
∴ Total = R1 900 + R684,00 = R2 584,00
(3)
20.
1557=105\tfrac{15}{57}=105/N (the tagged fraction in the sample matches the tagged fraction in the population)
N=105×57÷15N=105\times57\div15
N399N\approx399 impala
(3)
21.
Let the breadth =x=x, so the length =x+4=x+4
2(x+x+4)=282(x+x+4)=28
4x+8=284x+8=28
x=5x=5 (breadth 5 cm, length 9 cm)
∴ Area =9×5=45=9\times5=45 cm2^{2}
(5)
22.1
Mean =(25+33+31+28+33+45+29+24)÷8=(25+33+31+28+33+45+29+24)\div8
∴ Mean =248÷8=31=248\div8=31 goals
(2)
22.2
Ordered data: 24; 25; 28; 29; 31; 33; 33; 45
Middle two values: 29 and 31
∴ Median =(29+31)÷2=30=(29+31)\div2=30 goals
(2)
22.3
∴ Mode = 33 goals (it appears twice; every other value appears only once)
(1)
22.4
∴ Range = 45 − 24 = 21 goals
(1)
22.5
New mean =(24845)÷7=203÷7=29=(248-45)\div7=203\div7=29 goals
New median = 4th value of the remaining 7=297=29 goals
Mean: 31 → 29 (change 2); Median: 30 → 29 (change 1)
∴ The mean changes more (2 vs 1)
(4)
23.1
∴ 12 cm
(1)
23.2
Changes: Week 1→Week 2: −2; Week 2→Week 3: −4; Week 3→Week 4: −2; Week 4→Week 5: +4; Week 5→Week 6: +2
∴ Biggest increase: Week 4 → Week 5 (+4)
(2)
23.3
Week 1: 16 cm; Week 6: 14 cm
∴ Overall decreasing from Week 1 to Week 6
(2)
24.
Compare the two methods: a computer's random selection gives every tin in the batch an equal chance; the last tins packed may not represent the whole run.
∴ Method B
(2)
25.
Learners in at least one set = |Cricket| + |Hockey| − both = 19 + 20 − 8
= 31 learners
∴ Neither = 34 − 31 = 3 learners
(3)
26.1
S=S= {yellow; purple; blue; green}
(2)
26.2
n(S)=4n(S)=4
(1)
26.3
Each outcome is equally likely: P(one outcome) =14=\tfrac{1}{4}
Sum of all outcomes =4×14=44=4\times\tfrac{1}{4}=\tfrac{4}{4}
∴ The probabilities add up to 1
(2)
27.
Time in hours =150÷60=2,5h=150\div60=2{,}5h (60 min =1h=1h)
Distance =60×2,5=150=60\times2{,}5=150 km (distance = speed ×\times time)
∴ Petrol needed =150÷100×8=12=150\div100\times8=12 ℓ (the car uses 8 ℓ for every 100 km)
(4)
28.1
∴ Time in hours =180÷60=3h=180\div60=3h (60 min =1h=1h)
(2)
28.2
∴ Distance =80×3=240=80\times3=240 km (distance = speed ×\times time)
(2)
28.3
Petrol needed =240÷100×10=24=240\div100\times10=24 ℓ (the car uses 10 ℓ for every 100 km)
∴ Petrol cost =24×R21,95=R526,80=24\times\mathrm{R}21{,}95=\mathrm{R}526{,}80 (cost = litres ×\times price per litre)
(3)
29.
=1814411÷(27311)=18-\tfrac{144}{11}\div(27-\tfrac{3}{11}) (multiply straight across)
=1814411÷29411=18-\tfrac{144}{11}\div\tfrac{294}{11} (LCD = 11)
=182449=18-\tfrac{24}{49} (÷=×\div=\times reciprocal)
=172549=17\tfrac{25}{49} (LCD = 49)
(4)
30.
Total (after) = 190 + 14 = 204 (new sum = old sum + the new value)
∴ Mean (after) =204÷8=25,5=204\div8=25{,}5 minutes (mean = sum ÷\div count)
(3)
31.
the scale factor =6÷4=1,5=6\div4=1{,}5 (△RST ||| △XYZ)
YZ=18×1,5=27YZ=18\times1{,}5=27 cm (△RST ||| △XYZ)
(3)
32.
A(4; −6) → A′(−6; 4) (reflection in the line y=xy=x)
A′(−6; 4) → A″(6; −4): both coordinates change sign
∴ The second transformation is a rotation of 180° about the origin
(3)
33.
2x8=22x-8=-2 (subtract 3x from both sides)
2x=2+8=62x=-2+8=6 (do the same on both sides)
x=3x=3 (divide by the coefficient of x)
(3)
34.
Mistake: the legs were simply ADDED — Pythagoras squares them first.
h2=62+42=52h^{2}=6^{2}+4^{2}=52, so h=527,2h=\sqrt{52}\approx7{,}2 cm (not 10).
(2)
35.
Reason: solving gives x=1114=3x=11-14=-3, which is negative — not positive.
∴ Never true.
(2)
36.
✎ Teacher-marked — accept any mathematically sound answer covering the points below.
Model answer: a square metre is a square with BOTH sides 100 cm, so its area is 100×100=10000100\times100=10\,000 cm2^{2}. The length factor 100 must be used TWICE for an area.
1m2=100001m^{2}=10\,000 cm2^{2}.
(2)
37.1
Distance = speed ×\times time =100×1,5=100\times1{,}5
∴ = 150 km
(2)
37.2
150÷100×6150\div100\times6
∴ = 9 ℓ
(2)
37.3
9×R21,959\times\mathrm{R}21{,}95
∴ = R197,55
(2)
38.1
5x+27=3x+455x+27^\circ=3x+45^\circ (alt ∠s, AB ∥ CD)
2x=182x=18
x=9x=9
(3)
38.2
5(9)+27=725(9)+27=72^\circ
(1)
39.1
P=P= πr+2r=r+2r=×14\times14) + (2 × 14)
P43,98+28=71,98P\approx43{,}98+28=71{,}98 cm
(3)
39.2
A=A= ½πr2=r^{2}= ½ ×\times π ×142\times14^{2}
A307,88A\approx307{,}88 cm2^{2}
(2)
40.
x+76=118x+76^\circ=118^\circ (ext ∠ of △)
x=11876=42x=118^\circ-76^\circ=42^\circ
(2)
41.
Exactly one pair of opposite sides carries the parallel arrows; the other pair is not parallel.
∴ Trapezium
(2)
42.
Matched: DE=PQDE=PQ, EF=QREF=QR, DF=PRDF=PR
∴ △DEF ≡ △PQR (SSS)
(2)
43.
∴ ∠GEF=120÷2=60GEF=120^\circ{}\div2=60^\circ
(1)
44.
F+VE=2F+V-E=2
F=2+EV=2+2011F=2+E-V=2+20-11
F=11F=11 — it has 11 faces
(3)
45.1
A(−1; −3) → A′(2; −1)
B(−1; −1) → B′(2; 1)
∴ C(−5; −2) → C′(−2; 0)
(3)
45.2
∴ (x; y) → (x + 3; y + 2)
(2)
45.3
∴ Yes — congruent: all side lengths and angle sizes stay exactly the same
(1)
46.
∴ 8 sides
(1)
47.
Line of symmetry 1: from a corner to the midpoint of the opposite side.
Line of symmetry 2: from a corner to the midpoint of the opposite side.
Line of symmetry 3: from a corner to the midpoint of the opposite side.
Line of symmetry 4: from a corner to the midpoint of the opposite side.
Line of symmetry 5: from a corner to the midpoint of the opposite side.
Line of symmetry 6: from a corner to the midpoint of the opposite side.
∴ 7 lines of symmetry
(1)
48.
P(−2; 4)
∴ Q(−2; 3)
(2)
49.
Diameter =2×=2\times radius =2×6=2\times6
=12m=12m
(1)
50.
The two arms meet at M, so M is the vertex.
∴ ∠EMG (or equivalently ∠GME)
(1)
51.
370 cm =370÷100m=370\div100m
=3,7m=3{,}7m
(2)
52.1
SA=6s2SA=6s^{2} (a cube has 6 identical square faces)
=6×32=6×9=6\times3^{2}=6\times9
SA=54SA=54 cm2^{2}
(3)
52.2
V=s3=3×3×3V=s^{3}=3\times3\times3
V=27V=27 cm3^{3}
(2)
53.
ST2=RT2RS2ST^{2}=RT^{2}-RS^{2} (Pythagoras)
ST2=10282=10064=36ST^{2}=10^{2}-8^{2}=100-64=36
ST=36=6ST=\sqrt{36}=6 cm
(3)
54.
r=69r=69^\circ (∠s on a str line)
n=48n=48^\circ (vert opp ∠s)
z=63z=63^\circ (∠ sum of △)
(4)
55.
n=24n=24 cm (Pythagoras)
v=26v=26 cm (Pythagoras)
(3)
56.1
b=25b=25^\circ (vert opp ∠s)
(2)
56.2
m=76m=76^\circ (∠s on a str line)
(2)
56.3
r=104r=104^\circ (vert opp ∠s)
(2)
56.4
z=51z=51^\circ (∠s on a str line)
(2)
56.5
p=134p=134^\circ (∠s on a str line)
(2)