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7.2.3 Geometrical Constructions, PT3 Focus Practice

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Question 6:
Diagram below in the answer space shows part of a triangle ABC.
(a) Using a pair of compasses, protractor and ruler, draw the    triangle ABC with AB=7 cm, ∠ABC= 65 o  and BC=5 cm. (b) Measure ∠ACB.

Answer:
(a)



Solution:
(a)


(b)
∠ACB= 72 o



Question 7:
Diagram below shows a quadrilateral ABQR.


(a) Using only a ruler and a pair of compasses, construct the diagram using the measurement given.
Begin from the straight lines AB and BQ provided in the answer space.
(b) Based on the diagram constructed in (a), measure the length, in cm, of QR.

Answer:
(a)



Solution:
(a)


(b)
QR = 5.9 cm

7.2.2 Geometrical Constructions, PT3 Focus Practice

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7.2.2 Geometrical Constructions, PT3 Focus Practice 2

Question 4:
Diagram below shows a parallelogram PQRS. The point T lies on PQ such that ST is perpendicular to PQ.


 
Using only a ruler and a pair of compasses, construct parallelogram PQRS, beginning from the lines PQ and PS provided in the answer space.

Answer:
 


Solution:


Question 5:
Diagram below shows a triangle ABC.


 
(a) Using only a ruler and a pair of compasses, construct triangle ABC, beginning with the straight line BC in the answer space.
(b)   Hence, measure the length, in cm, of BA.

Answer:
(a)



Solution:
(a)

 
 
(b)
BA = 4.5 cm

7.2.1 Geometrical Constructions, PT3 Focus Practice

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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

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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

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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+( 25−5 )+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    =625−49    =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

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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,
BC2 = 52 – 42
= 25 – 16
= 9
BC = √9
  = 3 cm

In ∆ ABD,
BD = BC + CD
  = 3 + 6
  = 9
AD2 = 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,
AC2 = 62 + 82
= 36 + 64
= 100
AC = √100
  = 10 cm
AD = 5 cm

In ∆ EDC,
EC2 = 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,
BC2 = 252 – 242
= 625 – 576
= 49
BC = √49
  = 7 cm
AB = 17 – 7 = 10 cm

In ∆ DAB,
AD2 = 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
BC2 = 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:
AD2 = 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

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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.