7.2.1 Geometrical Constructions, PT3 Focus Practice


7.2.1 Geometrical Constructions, PT3 Focus Practice
Question 1:
Diagram below in the answer space shows part of triangle ABC.
(a) Using a pair of compasses, a protractor and a ruler, draw a triangle ABC starting from the line given in the answer space with AB = 4.5 cm, BC = 6 cm and ∠ABC = 105o.
(b) Measure ∠BCA.

Answer
:
(a)

 

Solution
:
(a)


(b)
∠BCA = 31o


Question 2:
Diagram below shows a triangle ABC drawn not to scale.

 
Diagram in the answer space shows a straight line AB.
(a) Using only a ruler and a pair of compasses, construct
 (i) triangle ABC to the measurements shown in diagram above,
 (ii)   the perpendicular line to the straight line AC which passes through the point B.
(b)   Measure the perpendicular distance between straight line AC and point B.

Answer
:
(a)(i),(ii)

 

Solution
:
(a)(i),(ii)

 (b)
5.3 cm


Question 3:
Diagram below shows a pentagon ABCDE.


Using only a ruler and a pair of compasses, construct the diagram, beginning from the straight lines CD and DE provided in the answer space.
 
Answer:

Solution:

 

5.4.3 Extraction of Metals from their Ores Using Coke


5.4.3 Extraction of Metals from their Ores Using Coke

1. In industries, metal ores which are less reactive than carbon are heated with carbon to obtain pure metal.
2. Pure metals which can be extracted using carbon are zinc, iron, tin and lead.

Extraction of Tin from its Ore


Extracting tin ore in a blast furnace
 
1. Tin ore exists naturally as cassiterite (or tin oxide).
2. Tin ore is washed with water to remove sand, clay and others impurities.
3. After that, tin ore is roasted to take away foreign matter such as carbon, sulphur and oil.
4. Lastly, the tin ore is mixed with carbon and limestone in the form of coal and is heated in a blast furnace at a high temperature.
5. The function of the limestone is to remove impurities.
6. Reduction reaction occurs during heating, carbon which is more reactive that tin removes oxygen from the tin oxide to produce pure tin and carbon dioxide.


7.
Pure tin flows out from the furnace into moulds to harden as tin ingots.
8. At the same time, the limestone (calcium carbonate) breaks down to form quicklime (calcium oxide) which reacts with impurities to form slag

6.2.2 Pythagoras’ Theorem, PT3 Focus Practice


Question 6:
In diagram below, ABCH is a square and DEFG is a rectangle. HCDE is a straight line and HC = CD.

Find the perimeter, in cm, of the whole diagram.

Solution:
G H 2 =D H 2 +D G 2    = 24 2 + 7 2    =576+49    =625 GH=25 cm Perimeter of the whole diagram =12+12+12+26+7+14+25 =108 cm

Question 7:
In the diagram, ABC and EFD are right-angled triangles.


Calculate the perimeter, in cm, of the shaded region.

Solution:
D E 2 = 3 2 + 4 2    =9+16    =25 DE= 25  =5 cm A C 2 = 7 2 + 24 2    =49+576    =625 AC= 625  =25 cm Perimeter of the shaded region =24+7+( 255 )+3+4 =58 cm

Question 8:
In the diagram, ABD and BCE are right-angled triangles. ABC and DBE are straight lines. The length of AB is twice the length of BC.


Calculate the length, in cm, of CE.

Solution:
A B 2 = 25 2 7 2    =62549    =576 AB= 576  =24 cm BC=24 cm÷2  =12 cm C E 2 = 5 2 + 12 2    =25+144    =169 CE= 169  =13 cm

Question 9:
Diagram below shows two right-angled triangles, ABE and CBD. BED is a straight line.
Find the length, in cm, of BC. Round off the answer to two decimal places.

Solution:
3 2 +B E 2 = 5 2    B E 2 = 5 2 3 2 =16 BE=4 cm B C 2 + ( 5+4 ) 2 = 17 2    B C 2 = 17 2 9 2 =208  BC= 208  BC=14.42 cm


Question 10:
Diagram below shows two right-angled triangles, ABC and ADE. ACD is a straight line.
Find the length, in cm, of AE. Round off the answer to one decimal places.

Solution:
A C 2 = 12 2 + 9 2    =225 AC= 225  =15 cm A E 2 = ( 15+9 ) 2 + 11.5 2    =576+132.25    =708.25 AE= 708.25  =26.6 cm

6.2.1 Pythagoras’ Theorem, PT3 Focus Practice


6.2.1 Pythagoras’ Theorem, PT3 Focus Practice
 
Question 1:
In the diagram, ABC is a right-angled triangle and BCD is a straight line. Calculate the length of AD, correct to two decimal places.

Solution:
In ∆ ABC,
BC= 52 – 42
= 25 – 16
= 9
BC = √9
  = 3 cm

In ∆ ABD,
BD = BC + CD
  = 3 + 6
  = 9
AD= 42 + 92
= 16 + 81
= 97
AD = √97
  = 9.849
  = 9.85 cm


Question 2:
In the diagram shows two right-angled triangles. Calculate the perimeter of the whole diagram.

Solution:
In ∆ ABC,
AC= 62 + 82
= 36 + 64
= 100
AC = √100
  = 10 cm
AD = 5 cm

In ∆ EDC,
EC= 122 + 52
= 144 + 25
= 169
EC = √169
  = 13 cm

Perimeter of the whole diagram
= AB + BC + CE + DE + AD
= 6 + 8 + 13 + 12 + 5
= 44 cm


Question 3:
Diagram below shows a triangle ACD and ABC is a straight line.
Calculate the length, in cm, of AD.

Solution:
In ∆ DBC,
BC= 252 – 242
= 625 – 576
= 49
BC = √49
  = 7 cm
AB = 17 – 7 = 10 cm

In ∆ DAB,
AD= 102 + 242
= 100 + 576
= 676
AD = √676
  = 26 cm


Question 4:
In diagram below ABDE is a square and EDC is a straight line.
The area of the square ABDE is 144 cm2.
Calculate the length, in cm, of BC.

Solution:
BD = √144
  = 12 cm
BC= 122 + 92
= 144 + 81
= 225
BC = √225
  = 15 cm


Question 5:
In the diagram, ABCD is a trapezium and AED is a right-angled triangle.
 
Calculate the area, in cm2, of the shaded region.

Solution:
AD= 52 + 122
= 25 + 144
= 169
AD = √169
  = 13 cm
Area of trapezium ABDC
= ½ (13 + 15) × 9
= 126 cm2

Area of triangle AED
= ½ × 5 × 12
= 30 cm2

Area of the shaded region
= 126 – 30
= 96 cm2

6.1 Pythagoras’ Theorem


6.1 Pythagoras’ Theorem

6.1.1 Pythagoras’ Theorem
1. In a right-angled triangle, the hypotenuse is the longest side of the triangle.
2. Pythagoras’ Theorem:

  In a right-angled triangle, the square
  of the hypotenuse is equal to the sum
  of the squares of the other two sides.
Example 1:
 
Solution:
x 2 = 5 2 + 12 2 = 25 + 144 x = 169 = 13  

Example 2:
 
Solution:
x 2 = 15 2 9 2 = 225 81 x = 144 = 12

3. Pythagorean triples
are three whole numbers that form the sides of a right-angled triangle.

Example:
(a) 3, 4, 5
(b) 6, 8, 10
(c) 5, 12, 13
(d) 8, 15, 17
(e) 9, 12, 15


6.1.2 The Converse of the Pythagoras’ Theorem
 
 
  In a triangle, if the sum of the squares of the two sides
  is equal to the square of the longest side, then the angle
  opposite the longest side is a right angle.

5.2.3 Ratio, Rates and Proportions (I), PT3 Focus Practice


Question 11:
Diagram below shows two baskets of eggs.

(a) State the ratio of the number of chicken eggs to the number of duck eggs.
(b) A number of eggs need to be added into each basket so that the ratio in (a) remain unchanged. Find the minimum number of eggs to be added in each basket.

Solution
(a)
Ratio of the number of chicken eggs to the number of duck eggs
= 12 : 16
= 3 : 4

(b)
Let the number of eggs need to be added into each basket = n
Total number of eggs need to be added into each basket
= 3n + 4n

Minimum value of n = 1
Hence, the minimum number of eggs to be added into each basket
= 3(1) + 4(1)
= 3 + 4
= 7 (3 chicken eggs and 4 duck eggs need to be added to each basket)  

 


Question 12:
Diagram below shows two baskets, A and B, filled with candies.
(a) State the ratio of the number of candies in basket A to the number of candies in basket B.
(b) If 20 candies are added into basket A, calculate the number of candies which need to be added into basket B so that the ratio in (a) remain unchanged.

Solution
(a)
Ratio of the number of candies in basket A to the number of candies in basket B
= 10 : 12
= 5 : 6

(b)
If 20 candies are added into basket A, the number of candies which need to be added into basket B so that the ratio 5 : 6 remain unchanged
= 20 ÷ 5 × 6
= 4 × 6
= 24

 


Question 13:
Karim built two towers using toy bricks of the same size as shown in Diagram below.

(a) State the ratio of the number of toy bricks in Tower P to the number of toy bricks in Tower Q.
(b) A number of toy bricks need to be removed from each tower so that the ratio in (a) remains unchanged.
Find the total number of toy bricks to be removed from each tower.

Solution
(a)
Tower P : Tower Q
= 12 : 18
= 2 : 3

(b)
Let number of toy bricks to be removed from each tower = n
Total number of bricks to be removed
= 2n + 3n
2(1) + 3(1) = 2 + 3 = 5 → (2 bricks from tower P, 3 bricks from tower Q)
2(2) + 3(2) = 4 + 6 = 10 → (4 bricks from tower P, 6 bricks from tower Q)
2(3) + 3(3) = 6 + 9 = 15 → (6 bricks from tower P, 9 bricks from tower Q)
2(4) + 3(4) = 8 + 12 = 20 → (8 bricks from tower P, 12 bricks from tower Q)
2(5) + 3(5) = 10 + 15 = 25 → (10 bricks from tower P, 15 bricks from tower Q)





5.2.2 Ratio, Rates and Proportions (I), PT3 Focus Practice


Question 6:
In a campaign of selling T shirt for Merdeka Day makes an amount of profit, RM x. The profit is divided between Jamal and Rafizi in the ratio of 5 : 3. Jamal receives RM 5400 more than Rafizi. Calculate the value of x.

Solution:
Portion of profit for Jamal = ⅝
Portion of profit for Rafizi = ⅜
Given Jamal receives RM 5400 more than Rafizi,
2 8 RM 5400 1 8 RM 5400÷2=RM2700 x=2700×8   =21600

Question 7:
Karim and Roger share RM300 in the ratio of 2 : 3. Karim gives ⅓ of his share to Mandeep. Then Mandeep received RM60 from Roger.
Find the ratio of Karim’s money to roger’s money to Mandeep’s money.

Solution:
Money received by Karim = 2 5 ×RM300 =RM120 Money received by Roger = 3 5 ×RM300 =RM180 Karim's money after given to Mandeep =RM120× 2 3 =RM80 Roger's money after given to Mandeep =RM180RM60 =RM120 Mandeep's money =RM40+RM60 =RM100 Ratio=80:120:100 =4:6:5

Question 8:
Liza received three types of coloured marbles, red, green and yellow in the ratio of 2 : 5 : x. Given that the number of yellow marbles are more than the number of red marbles but less than the number of green marbles.
Calculate the number of yellow marbles that Liza received if the total number of marbles is 120.

Solution:
red : green : yellow = 2 : 5 : x
Given 2 < x < 5, so possible value of x is 3 in order to get a round number from the total of 120 marbles.

yellow marbles total marbles = 3 2+5+3 yellow marbles 120 = 3 10 yellow marbles= 3 10 ×120 yellow marbles=36


Question 9:
John and Mahmud are required to draw a triangle ABC. The ratio of ∠A : ∠B : ∠C of the triangle drawn by John is 4 : 8 : 3 while the triangle drawn by Mahmud is 5 : 5 : 8.
Find the difference between the value of ∠B drawn by John and Mahmud.

Solution:
B drawn by John = 8 4+8+3 × 180 o = 96 o B drawn by Mahmud = 5 5+5+8 × 180 o = 50 o


Question 10:
The number of workers in an office building is 168 and they are placed in level two and level three. The number of workers in level three is 72.
(a) The ratio of workers in level two to level three is x : 3.
Find the value of x.
(b) 162 new workers have just started working in the office building. 3 of them are placed in level two, 93 in level three, while the rest is placed in level four.
Determine the ratio, in the lowest term of the workers of level two to three to four.

Solution:
(a) Ratio of workers in level two to level three=x:3 x:3=( 16872 ):72 x:3=96:72 x 3 = 96 72 x= 96 72 ×3 x=4

(b) Ratio of workers in level two to three to four =( 96+3 ):( 72+93 ):( 162393 ) =99:165:66 = 99 33 : 165 33 : 66 33 =3:5:2

5.2.1 Ratio, Rates and Proportions (I), PT3 Focus Practice


5.2.1 Ratio, Rates and Proportions (I), PT3 Focus Practice

Question 1:
Given x : y = 4 : 5 and x + y = 180. Find the value of x.

Solution:
Given x:y=4:5 and x+y=180 x x+y = 4 4+5 x 180 = 4 9 x= 4 9 ×180  =80
 

Question 2:
Given x : y = 9 : 5 and xy = 16. Find the value of x + y.

Solution:
Given x:y=9:5 and xy=16 x+y xy = 9+5 95 x+y 16 = 14 4 x+y= 7 2 ×16 =56 x+y=56



Question 3:
Amy, Rizal and Muthu donated to a charity fund in the ratio of 2 : 1 : 3. The total donation from Amy and Muthu was RM360.
Calculate the amount donated by Rizal.

Solution:
Amy : Rizal : Muthu = 2 : 1 : 3
Rizal Amy + Muthu = 1 2+3 Rizal RM360 = 1 5 Rizal= 1 5 ×RM360 =RM72



Question 4:
In the figure, TU: UV : VW = 5 : 7 : 2.
 
If TU = 30 cm, what is the length of TW?

Solution:
Given TU:UV:VW=5:7:2 TU=30cm TU+UV+VW TU = 5+7+2 5 TW 30 = 14 5 TW= 14 5 ×30 =84cm


Question 5:
In diagram below, ABC is a triangle.
The sides of the triangle are in the ratio AB : BC: CA = 4 : 7 : 6.
Find the difference in length, in cm, between AB and CA.

Solution:
AB:BC:CA=4:7:6 BC=35cm AB BC = 4 7 AB 35 = 4 7 AB= 4 7 ×35 =20cm CA 35 = 6 7 CA= 6 7 ×35 =30cm

Difference in length between AB and CA
= 30 – 20
= 10 cm

3.4.1 The Effects of Environmental Pollution (Structured Questions)


Question 1:
Diagram 1.1 shows a type of an environmental pollution in an industrial area.


(a)(i) Based on Diagram 1.1, what is released by the factory? [1 mark]

(a)(ii) State the type of pollution based on the answer in (a)(i). [1 mark]

(b) State the effect of pollution shown in Diagram 1.1 on human being. [1 mark]

(c) Suggest one method to control the pollution shown in Diagram 1.1. [1 mark]


(d) Diagram 1.2 shows the effect of environmental pollution due to the heat trapped by gas P.

(i) Name gas P. [1 mark]
(ii) Name the phenomenon shown in Diagram 1.2. [1 mark]


Answer:
(a)(i)
Thick black smoke (soot)

(a)(ii) Air pollution

(b)
The acidic gases in the smoke irritate and corrode the respiratory passage (causing asthma and bronchitis)

(c)
Fit electrostatic precipitators on the chimneys to attract the soot to the inner wall of the chimneys.

(d)(i)
Carbon dioxide

(d)(ii)
Global warming/ Greenhouse effect


Question 2:
Diagram 2 shows the atmospheric layer around the Earth.


(a)(i) What is layer K? [1 mark]

(ii) State the importance of layer K. [1 mark]

(b)(i) Name the substance which can damage layer K. [1 mark]

(ii) Name one appliance that releases the substance in (b)(i). [1 mark]

(c) State two effects if layer K is damaged. [2 marks]


Answer:
(a)(i)
Ozone layer

(a)(ii)
Absorbs harmful ultraviolet rays in sunlight

(b)(i)
Chlorofluorocarbon (CFC)

(b)(ii)
Aerosol spray

(c)
1. Ultraviolet ray reaching the Earth can cause skin cancer and cataract
2. Ultraviolet ray lower our body’s immunity to diseases.


4.2.1 Linear Equations I, PT3 Practice 1


4.2.1 Linear Equations I, PT3 Practice 1
Question 1:
Solve the following linear equations.
  (a)  4 – 3n = 5n – 4   
  (b) 4 m 2 10 + m = 1 2  

Solution:
(a)
4 – 3n = 5n – 4   
–3n –5n = – 4 – 4   
–8n = – 8   
 8n = 8
n = 8/8 = 1
 
(b) 4 m 2 10 + m = 1 2 2 ( 4 m 2 ) = 10 + m 8 m 4 = 10 + m 8 m m = 10 + 4 7 m = 14 m = 14 7 m = 2


Question 2:
Solve the following linear equations.
(a) x 3 = 4 x (b) 3 ( x 2 ) 5 = 9
 
Solution:
(a) x 3 = 4 x x = 3 ( 4 x ) x = 12 3 x x + 3 x = 12 4 x = 12 x = 12 4 x = 3

(b) 3 ( x 2 ) 5 = 9 3 ( x 2 ) = 9 × 5 3 x 6 = 45 3 x = 45 + 6 3 x = 51 x = 51 3 x = 17



Question 3:
Solve the following linear equations.
(a) 3 m 4 + 15 = 9 (b) 2 m + 8 = 3 ( m 2 )

Solution:
(a) 3 m 4 + 15 = 9 3 m 4 = 9 15 3 m 4 = 6 3 m = 6 × 4 3 m = 24 m = 24 3 m = 8
 
(b) 2 m + 8 = 3 ( m 2 ) 2 m + 8 = 3 m 6 2 m 3 m = 6 8 m = 14 m = 14


Question 4:
Solve the following linear equations.
(a) 11 + 2 x 3 = 9 (b) x 5 3 = x 6

Solution:
(a) 11 + 2 x 3 = 9 2 x 3 = 9 11 2 x 3 = 2 2 x = 6 x = 6 2 x = 3
 
(b) x 5 3 = x 6 6 ( x 5 ) = 3 x 6 x 30 = 3 x 6 x 3 x = 30 3 x = 30 x = 10


Question 5:
Solve the following linear equations.
(a) 4 ( 2 x 3 ) = 24 (b) y 2 y + 4 3 = 5

Solution:

(a) 4 ( 2 x 3 ) = 24 8 x 12 = 24 8 x = 24 + 12 8 x = 36 x = 36 8 x = 4 1 2
 
(b) y 2 y + 4 3 = 5 y × 3 2 × 3 2 ( y + 4 ) 3 × 2 = 5 3 y 2 ( y + 4 ) 6 = 5 3 y 2 y 8 = 30 y = 30 + 8 y = 38