Additional Mathematics 3

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Additional Mathematics

3/2011Smk Sungai Pusu

Nur Hafiza Binti AbuKassim

5 Avicenna940816-14-5254

Miss Noor AishahHamdan


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Apply mathematics to everyday situation and appreciate the

importance and the beauty of mathematics in everyday lives

To improve problem-solving skills, thinking skills, reasoning and

mathematical communication.

To develop positive attitude and personalities and intrinsic

mathematical values such as accuracy, confidence and systematic

reasoning.

To stimulate learning environment that enhances effective learning,

inquiry-based and our team work.

To develop mathematical knowledge in a way which increase our

interest and confidence.

Section 1


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Section 2Method 1

a)1.Divide area p into 3 segments.2 triangles and 1 rectangle

Triangle 1: 2 1

= 1 m

Triangle 2: 4 3

= 6 m


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Rectangle 1: 2 4

= 8 m

P = (1+6+8) m

P = 15 m

2.Divide area Q into 2 segments,1 triangle and 1 rectangle

Triangle 1 : 4 3

= 6 m

Rectangle 1: 2 3

=6 m

Q=(6 + 6)m

Q=12 m

3.Divide area R into 3 segments.1 triangles and 2 rectangles

Triangle 1: 1 2

= 1 m

Rectangle 1: 2 3

= 6 m

Rectangle 2: 4 2

= 8 m


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R = (1+6+8) m

R =15 m

Method 2

Coordinate geometry method:


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077430

002240

(0 + 14 + 14 + 16 + 0 ) (0 + 0 + 8 + 6 + 0 )

= (30)

=15 m

Area Q


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03300

04660

= (0 + 18 + 18 + 0 )(0 + 12 + 0 + 0 )

= (24)

= 12 m

Method 3

Integration methodArea P:


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mAE = 40

30

mAE = 4

3

Equation AE

Y = 43x

mED = 2443

mED=2


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Equation ED

Y=2x + 10

[034 3dx + 34][2x +10 dx + 472dx]

= 2x2330 + (-x2) + 1043 + 2274

= 60 + 2421 + 148

= 6 + 3 + 6

= 15 m

Area Q

mAE = 43


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Equation AE

y = 4x3

x = 3y4

= 043y4dy + 463 dy

= 3y2840 + 3y64

= 6 + 6

= 12 m

Area R

076 dy 12 15

=6x70 27

=420 27

=4227

=15 m

Verification answer by Geo Gebra :

Area P


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Area Q

Area R


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b)The mathematics society wishes to fence up the remainingsides of the region P. Determine the length of the fencerequired:

from the diagram:

AE2 = (62)2 + 32

AE2 = 42 + 32

AE2 = 25

AE = 25

AE = 5 m

AE2 = [7(3+3)2] + (42)2

AE2 = 12 + 22

AE2 = 5

AE = 2.236 m

CD = 3 m

BC = (6 4)

BC = 2 m

AB = 7 m

AEDCB = 2 + 3 + 2.236 +5 m

AEDCB = 12.236 m


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c) It is impossible for the society to carry out the fencing with an

allocation of RM250.00.

This is because:

RM250.00 + RM25/m

= 10m

Therefore, RM250.00 can only cover up to 10m of length

Length of AEDCB is 12.236m

= 12.236 RM25.00

= RM305.90 is needed to carry out the fencing plan.

d. i) 1 point for the flag chain to be tied at E

1 point for the flag chain to be tied at a point along the hedge AB

Therefore: 2 points

ii) maximum area of the triangle obtained:

= 9.20 m AE

= 9.20 m 5 m

= 4.20 m

Then divide the triangle obtain into 2 right triangle:

Triangle 1


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Calculate the length.

Using Pythagoras theorem.

L = 52 (62) 2

L = 9

L = 3 m

Area =12 3 4

Area = 6 m

Triangle 2

Calculate the length.

Using Pythagoras theorem.

L = 4.22622

L = 1.64

L = 1.281 m

Area = 12 4 1.281

Area = 2.562 m

Maximum area of triangle:

= (6 + 2.562 ) m

= 8.562 m

Solution of triangle method:

12 ab sin C

Angle EAL tan-143=53.13

= 12 [5 ( 3 + 1.282 )( sin53.13 )]


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= 8.562 m

Section 3

Calculating angle AED by using 2 methods:

Method 1:

Draw a horizontal and vertical lines on Geogebra


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The angle on the left and rightside of AED is calculated:

AED = 180 (53.13 63.43)

AED = 63.44


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Geo Gebras calculations:

So,AED = 63.43

Method 2

Using the formula from the solution of triangle.

a2 = b2 + c2 2bc A

202 = 52 + 52255 A

20 = 5 + 25 255 A

10 = 255 A

10255 = A

0.4472 = A

A = 63.44


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a. Volume of water that has to be pumped in to fill up 80% ofthe pond.

So, the depth of the pond is 1m.

=12r 2

=122.236263.43 3.142180

=122.23621.017rad


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=2.767m 1 m

=2.767m 80%

=2.214m

b. i) The rate of change of depth of the water

dhdt = dhdv dvdt

dh

dvdt = 0.001m3s-1

dhdt =12.767 0.001

dhdt = 0.000361ms-1

ii) The depth of water after 10 minutes

= 0.00036ms-1 60s 10min

= 0.2166m

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

So the water maximum depth of the pond is 1m.Ratio:

0.2166m = 10min

1m = ??????min


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0.21661 = 10x

0.2166x = 10

x = 100.2166

x = 46.17min / 46min10s

iv) the minimum time taken, in minutes, before the wateroverflows,if the

pond is triangular shaped AED and has a depth of 2 metres.

Now the pond is triangular shaped. Solution of triangle formulacan be applied.

A = 12ab C

A = 12(5)(2.236)(63.43)

A = 5m

V=5 m 2m

V=10m

dhdt = dhdv dvdt

dhdt = 210 0.001

dhdt = 0.00021ms-1

So ratio might work.

Height of water after 10 mins:

= 0.0002ms-1 60s 10min

= 0.12 m


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Then using ratio,

0.12 m = 10min

2m = ?????min

0.122 = 10z

0.12z = 20

z = 200,12

z = 166.67min / 166min40s

Further exploration

a. i)the map in diagram 3


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since the scale is 60km per square,we may find thedistancebetween x and the city of Malacca by:

d = 60 km 6 square

d = 360km

ii)the formula given:

distance = nautical miles

1 nautical mile = 1.852 kilometres

= difference in latitudes in degrees

Malaccas latitude =217'N

Hence,

d = 60 nautical miles

d = 541'N 217'N 60(1.852)

d = 324' 111.12

d =377.8 km

Differences between the answers.

Yes, there is a different between the answers obtained. This isbecause the calculation by using the scale given by a map is onlyan approximation method. The answer is correct but lessaccurate compared to the answer from the calculation based onthe formula given. By using the latitudes, the answer is veryaccurate and significant.


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Additional Mathematics 3

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