Add Maths Task 3 2011

8/6/2019 Add Maths Task 3 2011

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www.krishnandoreme.co.cc 201

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Junior High math students can try a quiz that practices plotting points on such a grid.

Descartes invented the system of using the first letters of the alphabet to represent known

quantities, and the last letters to represent unknowns. For example, in the quadraticequation ax2 + bx + c = 0, the letters a, b, and c represent known values (coefficients),while x represents the unknown solution to the equation.

He also introduced the method for writing powers that we are familiar with. As anexample, if you want to write 4 x 4 x 4, you can use 43.

Descartes also investigated Euler's relation v + f - e = 2,the equation that describes how the number of verticies,

faces, and edges of a convex polyhedron are related.

In the example at the left, the prism has 8 faces, 12

verticies, and 18 edges. The rule states that, just as forthis figure where 8 + 12 - 18 = 2, the three values must

always equal 2, for any convex polyhedron.

In the area of science, Descartes did work in optics, astronomy and meteorology, but his

work here was flawed, and nowhere near as impressive as his results in mathematics andphilosophy.

His most famous philosophical work dealt with the concept of 'existence'. In hisinvestigation of reality, Descartes decided his first step would be to discover some fact

that was indisputable. He realized that there was one thing he never doubted that he

existed. "I think," he wrote, "therefore I am." (in Latin: cogito, ergo sum). It didn't matter

to Descartes whether this thinking was part of a dream or a hallucination, or even if hewas crazy. The fact that thought was going on proved that he existed, because there had

to be a thinker.

Having established that he existed, Descartes advanced his arguments for the

existence of God. The first of these was that he had the idea of a Perfect Being in his

imperfect mind. But he reasoned that an imperfect mind couldn't come up with the ideaof a Perfect Being, so there must actually be a Perfect Being (God) who gave him the

idea!

The second argument was this: If God is perfect -- as we imagine Him to be --

then He must exist, because if he didn't exist, He wouldn't be perfect. (If this argumentstrikes you as a bit strange, you are not alone. Most philosophers regard it as more of a

play on words than as a philosophical proof.)

From his conclusion that a perfect God exists, Descartes argued that God would

not deceive his created beings, so the things we experience around us must also be real.While other philosophers had argued that God exists by saying the universe must have a
http://www.worsleyschool.net/science/files/plot/plot.htmlhttp://www.worsleyschool.net/science/files/euler/euler.htmlhttp://www.worsleyschool.net/science/files/plot/plot.htmlhttp://www.worsleyschool.net/science/files/euler/euler.html


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creator, Descartes took the argument the other way. The world must exist, he argued,

because God exists.

Descartes believed that all material bodies, including the human body, aremachines that operate by mechanical principles. In his physiological studies, he dissectedanimal bodies to show how their parts move. He argued that, because animals have no

souls, they do not think or feel; thus vivisection, which Descartes pioneered, is

permissible.

In 1649 Queen Christina of Sweden persuaded Descartes to go to Stockholm.

However, the Queen, being interested in mathematics, wanted to draw tangents at 5 a.m.,and Descartes broke the habit of his lifetime of getting up at 11 o'clock. After only a few

months in the cold northern climate, walking to the palace for 5 o'clock every morning,

he died of pneumonia.

Part 2

Malaysia with its warm tropical climate is rich in flora and fauna. Beautiful gardens

are found all over Malaysia. SMK Permata decided to beautify the school compound by

getting the students involved in the planting and maintenance of the greenery in the school

compound as shown in Diagram 1. Each society is allocated a plot of land in various shapesand sizes to nurture throughout the year. The Mathematics Society, English Language

Society and Malay Language Society are allocated the region P, Q and R respectively asshown in Diagram 1.


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(a) Determine the area of regionP, Q andRby using at least three different methodsincluding the use of calculus. Verify the answers obtained by using computer

software. (Suggestions: GeoGebra, GSP, graphing calculator etc).

ANSWER PART 2(a)

(i) Area of regionPMethod 1

)32()1)(42(2

134

2

1++

++

=

= 6 + 3 + 6

= 15 cm2

Method 2

0

0

0

7

2

7

2

4

4

3

0

0

2

1=

( ) ( ))0(0)7(2)7(2)4(4)3(0)0(7)0(7)2(4)2(3)4(02

1 ++++++++=

44142

1=

312

1=

)30(2

1=

= 15 cm2


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Method 3

Gradient, m of lineAE,03

04

=

3

4=

At point (3, 4), m3

4= . Therefore the equation of straight lineAEisy x

3

4= .

Gradient, m of lineED,43

24

= 2=

At point (3, 4), m 2= . Therefore the equation of straight lineED is 102+=

xy.

Since line DC is horizontal, then gradient, m of lineDC= 0

At point (7, 2), m 0= . Therefore the equation of straight lineDCis 2=y .

+++

=

7

4

4

3

3

0

2)102(3

4dxdxxdxx

[ ]74

4

3

23

0

2

210

2

2

)2(3

4xx

xx +

+

+

=

[ ] [ ])4(2)7(2))3(10)3(())4(10)4((06

)3(4 222

++++

=

( ) ( )[ ] 630940166

36++++

=

= 6 + (24 21) + 6= 12 + 3= 15 cm2

(ii) Area of region QMethod 1

)32(342

1 +

=

= 6 + 6

= 12 cm2


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Method 2

0

0

4

3

6

3

6

0

0

0

2

1=

( ) ( ))0(4)3(6)3(6)0(0)0(3)4(3)6(0)6(02

1 ++++++=

361221 =

242

1=

)24(2

1=

= 12 cm2

Method 3

Since line HG is horizontal, then gradient, m of lineHG = 0

At point (3, 6), m 0= . Therefore the equation of straight lineHG is 6=y .

= 3421

63

0

dx

[ ] 66 30 = x= [ ]0)3(6 6

= 18 6= 12 cm2


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(iii) Area of regionR

Method 1

)42(2)43(2

1 +

+=

= 7 + 8

= 15 cm2

Method 2

6

3

4

3

2

4

2

7

6

7

6

3

2

1=

( ) ( )[ ])3(4)3(2)4(2)7(6)7(6)6(3)4(4)2(7)2(7)6(32

1++++++=+=

110802

1=

302

1=

)30(2

1=

= 15 cm2Method 3

Since line GF is horizontal, then gradient, m of line GF= 0

At point (7, 6), m 0= . Therefore the equation of straight linegfis 6=y .

+

+= )32(1)42(21

6

7

3

dx

[ ] )63(6 73 += x= [ ])3(6)7(6 9= 24 9= 15 cm2


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(b) Suppose there is a hedge alongAB. The Mathematics Society wishes to fence up the

remaining sides of the regionP. Determine the length of fence required.

ANSWER PART 2(b)

Perimeter of regionP=AE + ED + DC + CB + AB

AE= 22 43 +169+=

25=

= 5 m

ED = 22 12 +5=

= 2.236 m

DC= 3 m

CB = 6 4

= 2 m

AB = 7 m

Therefore, the perimeter of region P= 5 + 2.236 + 3 + 2 +7

= 19.236 m


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Since there is hedge alongAB, the remaining sides of regionPthat need to be fence up is

= 19.236 7

= 12.236 m

Therefore the length of the fence required is 12.236 m

(c) If a meter of fence costs RM25.00, what is the total cost required by the MathematicsSociety to fence up region P? Is it possible for the society to carry out the fencingwith an allocation of RM250.00? Explain your answer.

ANSWER PART 2(c)

1 m of fence = RM 25.00

Total cost required by the Mathematics Society to fence up regionPis

= 00.25236.12 = RM 305.90

Therefore, it is not possible for the mathematics Society to carry out the fencing with

an allocation of RM 250.00. It is because 1 m of the fence cost RM 25.00. The total

length of remaining sides of regionPthat has to be fence up is 12.236 m. So the totalamount will be RM 305.90. This value is more than RM 250.00. So, the allocation

not enough and the society still need RM 55.90 to carry out the fencing of remainingsides of region P.

(d) During the Mathematics Week, the society was given a single flag chain of length

9.20 meters to be used completely. The President of the society wishes to tie the flagchain continuously from A to E and then to another point along the hedge AB to

create a triangular-shaped area.(i) Make a conjecture about the number of points that the flag chain can be tied

to alongAB.

Support your conjecture with suitable calculations. Explain your answer.

(ii) Calculate the maximum area of the triangle obtained. Discuss.


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ANSWER PART 2( d)(i)

Plot 4.2 m line from pointEto lineAB

Find the angleEAB

3

4tan 1=EAB

= 53.13

It is a acute angle. Therefore by using ambiguous case, there will be another

triangle that can be formed. So, it will be two triangles that are formed. Therefore, theconjecture of point that can be tied along lineAB is 2.

4.2 m

4.2 m

4m

3m

4.2 m4.2m

3m

53.13


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ANSWER PART 2(d)(ii)

The maximum area of the triangle obtained,

5

sin

2.4

13.53sin AGE=

5

sin1905.0

AGE=

AGE= sin51905.0 AGE=sin9524.0

AGE= 9524.0sin 1

= 25.72AGE 25.72180 =EVA

= 107.75

= 13.5375.107180AEV= 19.12

=25.7225.72180VEG

= 35.50

Area of triangleAVE

)12.19sin(2.452

1=

439.3= m2

4.2 m4.2m

107.75

53.13

35.5019.12

72.25


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Area of triangle VEG

)50.35sin(2.42.42

1=

122.5=

m2

By comparing the area of triangleAVEand VEG, the maximum area of triangle obtained is

5.122 m2 that is triangle VEG.

Part 3

The Mathematics society decided to build a pond in regionPas shown in Diagram 2. The

pond is in the shape of a sector with centreE, radiusED and a depth of 1 meter.


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(a) Calculate the angleAED, in radians, by using at least two different methods.

ANSWER PART 3(a)

Method 1:-

Let angle AED is equal to y x+o o

Where xo represents the angle of triangular-shape at the right side inside the pond

The opposite side = 1 m

The adjacent side = 2 m

otan xa

=o

Therefore,1

2tan x =

o

1 1

2 x tan

=o

26.57x =o o

While yo

represents the angle of triangular-shape which is opposite the triangle with the angle xo

The opposite side = 3 m

The adjacent side = 4 m

otan y

a=

o

Therefore,3

4tan y =

o

1 3

4 y tan

=o

36.87y =o o

Note that angle AED is equal to y x+o o

Therefore, angle AED = 26.57 36.87+o o

= 63.43 o

In radian, 63.43180

o

o


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Method 2:-

Area of triangle =

Let the angle AED equal to

is the angle of triangular-shape at right side inside the pond

Area of triangular-shape at right side inside the pond

=

Therefore,

Therefore, angle of

In radian,

=

is the angle of triangular-shape at the left side inside the pond

Area of triangular-shape which is opposite the triangle with the angle

=

=

Therefore,

Therefore, angle of

In radian,

=

Note that angle AED is equal to

Therefore,

=


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(b) Determine the volume of water that has to be pumped in to fill up 80% of the pond.

ANSWER PART 3(b)

Volume =21

2r h

ED = 2 22 1+

= 5

= 5m

Note that the radius of the pond is 5m

Therefore, volume of the pond is equal to

( )21 5 1.107 1

2

= ( )2.5 1.107

= 32.768 m

The volume of water that has to be pumped to fill up 80% of the pond is


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802.768

100

= 32.214 m

(c) If the water is pumped into the pond at a constant rate of 0.001 m3s-1, calculate

(i) the rate of change of depth of the water,

ANSWER PART 3(c)(i)

Rate of change of depth the water isdh dh dV

dt dV dt =

Given that3 1

0.001dV

m s

dt

=

21

2V r h=

21

2

dVr

dh=

( )21

5 1.1072

=

2.768=

Therefore,1

2.768

dh

dV=

10.001

2.768

dh

dV=

0.36127 0.001=

4 3 1

3.6127 10 m s =

Therefore, the rate of change of depth of the water is

= 4 3 13.6127 10 m s

(ii) the depth of water after 10 minutes,

ANSWER PART 3(c)(ii)

The depth of the water after 10 minutes

( )43.6127 10 10 60= 4

3.6127 10 600=


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= 0.2168 m

(iii) the minimum time taken, in minutes, before the water overflows, and

ANSWER PART 3(c)(iii)

Note that 32.768V m=3 1

0.001dV

m sdt

=

The minimum time taken, in minutes, before the water overflows is

60V

dV

dt

=2.768

600.001

= 2768 60

= 46.13

= 46 minutes


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(iv) the minimum time taken, in minutes, before the water overflows, if the pondis triangular-shapedAED and has a depth of 2 meters.

ANSWER PART 3(c)(iv)

Volume of triangular-shaped AED pond =1

sin2

ab C h

Note that the height of the pond is 2 m

The angle of AED is 63.43 o

ED = 5 m

AE= 5 m

Therefore, the volume of triangular-shaped AED pond is

=1

5 5 sin 63.43 22

o

= 5 2

= 310 m

3

10V m=

3 10.001

dVm s

dt

=

Therefore, the minimum time taken, in minutes, before the water

overflows,

60V

dV

dt

=10

600.001

= 10 000 60

= 166.7

= 167 minutes


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FURTHER EXPLORATION

Maps have been used for thousands of years to aid travelers during their journey from

one place to another. Maps can also be used to estimate distance between places. In theyear 2014, a recreation park will be constructed in town marked X on the map of

Malaysia in as shown Diagram 3. This town has the latitude of 541 N and has the same

longitude as the city of Malacca.


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Explore and find the distance between these two places in kilometer by using,

(i) the map in Diagram 3ANSWER

Since the scale is 60 km per square, we can approximately find the distance X and the

city of Malacca by

60=d km 6 squaresd= 360 km

(ii) the formula given below:

Where difference in latitudes in degrees. Is there a difference between the answersobtained? Explain.

ANSWER

1 nautical mile = 1.852 kilometres.

= difference in latitudes in degrees.

Malaccas latitude = 217N

Thus,

60=d nautical miles)852.1(60)'172'415( = NNd

12.111'243( =d

8.377=d km

Yes, there is a different between the answer obtained. This is because the

calculation by using the scale given by a map is just an approxiamation method.

The answer is correct but less accurate compared to the answer from thecalculation based on the formula given. By using the latitudes, the answer is very

accurate and significant.


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A. Hospital Mesra Bukit Padang

JalanKolam, Kota Kinabalu, Sabah

E. PusatPakarPerubatan Kota Kinabalu

LorongMargosa, Taman LuyangFasa 8, Kota Kinabalu, Sabah

(i) Solve the triangle obtained,

Refer map 1.(ii) The shortest distance from your school to the line joining the two hospitals/clinics,

Based on map 1,the shortest distance from my school to the line joining the two

hospitals/clinics is 0.859km.


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REFLECTION

In doing this project, there are some values that I have practiced as cooperation in group.I have worked together with my friends from my school and also several friends from

other school.

Other than that, I have learned about the meaning of patience during worked up for this

assignment. I also have learned how to persevere in doing the job.

Moreover, I was also adopted in cooperation it other partners to complete this work.

Without help from them, I will not be able to complete this task.

I would be a hard working and trying to finish this job properly and just in time.

Next, I also learned to be a person who does not easily give up in doing any work.

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Add Maths Task 3 2011