# Form 5 Add Math Project 2009

## Form 5 Add Math Project 2009

**PART 1**There are a lot of things around us related to circles or parts of a circle.

(a)

Collect pictures of 5 such objects. You may use camera to take pictures

around your school compound or get pictures from magazines, newspapers,

the internet or any other resources.

(b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π.

__PART 2__(a) Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and BCR of diameter d

_{1}and d

_{2}respectively are inscribed in the semicircle PQR such that the sum of d

_{1}and d

_{2}is equal to 10 cm.

Complete Table 1 by using various values of d

_{1}and the corresponding values of d

_{2}. Hence, determine the relation between the lengths of arcs PQR, PAB and BCR.

(b) Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB, BCD and DER of diameter d

_{1}, d

_{2}and d

_{3}respectively are inscribed in the semicircle PQR such that the sum of d

_{1}, d

_{2}and d

_{3}is equal to 10 cm.

i) Using various values of d

_{1}and d

_{2}and the

corresponding values of d3, determine the relation between the lengths

of arcs PQR, PAB, BCD and DER. Tabulate your findings.

(ii)

Based on your findings in (a) and (b), make generalisations about the

length of the arc of the outer semicircle and the lengths of arcs of

the inner semicircles for n inner semicircles where n = 2, 3, 4....

(c) For different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.

__PART 3__The

Mathematics Society is given a task to design a garden to beautify the

school by using the design as shown in Diagram 3. The shaded region

will be planted with flowers and the two inner semicircles are fish

ponds.

(a) The area of the flower plot is y m

^{2}and the diameter of one of the fish ponds is x m. Express y in terms of it and x.

(b) Find the diameters of the two fish ponds if the area of the flower plot is 16.5 m

^{2}. (Use π=22/7 )

(c)

Reduce the non-linear equation obtained in (a) to simple linear form

and hence, plot a straight line graph. Using the straight line graph,

determine the area of the flower plot if the diameter of one of the

fish ponds is 4.5 m.

(d) The cost of constructing the fish ponds

is higher than that of the flower plot. Use two methods to determine

the area of the flower plot such that the cost of constructing the

garden is minimum.

(e) The principal suggested an additional of

12 semicircular flower beds to the design submitted by the Mathematics

Society as shown in Diagram 4. The sum of the diameters of the

semicircular flower beds is 10 m.

The diameter of the smallest flower bed is 30 cm and the diameter of

the flower beds are increased by a constant value successively.

Determine the diameter of the remaining flower beds.

由Giraffe于周四 九月 24, 2009 7:09 am进行了最后一次编辑，总共编辑了3次

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## 回复： Form 5 Add Math Project 2009

Well, this year's form 5 add math project is really easy. Much easier

than previous years. Therefore, you are encouraged to do it by

yourself, and then compare your answer with the proposed solution here,

or discuss your problem with our friends here.

The proposed

solution is NOT mean for you to copy. It just serve as a reference,

beside being a platform for all students to discuss their work.

Well, I leave this to your own self. There thousands of object which is round in shape.

There

are many websites providing information of pi. So, just google it.

Below are some websites I found through google. I think it's more than

enough for you to write a thesis.

Definition: Ratio of circle's circumference to its diameter

http://video.google.com/videoplay?docid ... 4291031420

http://en.wikipedia.org/wiki/Pi

http://reference.allrefer.com/encyclopedia/P/pi.html

http://www.bartleby.com/65/pi/pi.html

http://www.britannica.com/EBchecked/topic/458986/pi

http://encarta.msn.com/encyclopedia_761552884/Pi.html

http://scienceblogs.com/goodmath/2006/08/post_2.php

http://mathforum.org/dr.math/faq/faq.pi.html

http://www.gap-system.org/~history/Hist ... _ages.html

The length of arc (s) of a circle can be found by using the formula

where r is the radius.

From the table, we can conclude that

Length of arc PQR = Length of arc PAB + Length of arc BCR

Again, we use the same formula to find the length of arc of PQR, PAB, BCD and DER.

This time, the table is a big one. You can click on the table to change to the full size view.

Again, we can conclude that:

Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of arc CDR

Base on the findings in the table in(a) and (b) above, we conclude that:

The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles.

Diagram above shows a big semicircle with n number of small inner circle. From the diagram, we can see that

The length of arc of the outer semicircle

The sum of the length of arcs of the inner semicircles

Factorise π/2

Substitute

We get,

where d is any positive real number.

We can see that

As a result, we can conclude that

The

length of the arc of the outer semicircle is equal to the sum of the

length of arcs of any number of the inner semicircles. This is true for

any value of the diameter of the semicircle.

In

other words, for different values of diameters of the outer semicircle,

show that the generalisations stated in b (ii) is still true.

than previous years. Therefore, you are encouraged to do it by

yourself, and then compare your answer with the proposed solution here,

or discuss your problem with our friends here.

The proposed

solution is NOT mean for you to copy. It just serve as a reference,

beside being a platform for all students to discuss their work.

**PART 1***(a)*

Collect pictures of 5 such objects. You may use camera to take pictures

around your school compound or get pictures from magazines, newspapers,

the internet or any other resources.Collect pictures of 5 such objects. You may use camera to take pictures

around your school compound or get pictures from magazines, newspapers,

the internet or any other resources.

Well, I leave this to your own self. There thousands of object which is round in shape.

*(b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π.*There

are many websites providing information of pi. So, just google it.

Below are some websites I found through google. I think it's more than

enough for you to write a thesis.

Definition: Ratio of circle's circumference to its diameter

__Video__http://video.google.com/videoplay?docid ... 4291031420

__Encyclopedia__http://en.wikipedia.org/wiki/Pi

http://reference.allrefer.com/encyclopedia/P/pi.html

http://www.bartleby.com/65/pi/pi.html

http://www.britannica.com/EBchecked/topic/458986/pi

http://encarta.msn.com/encyclopedia_761552884/Pi.html

__Webpage__http://scienceblogs.com/goodmath/2006/08/post_2.php

http://mathforum.org/dr.math/faq/faq.pi.html

http://www.gap-system.org/~history/Hist ... _ages.html

__PART 2__*(a)*

Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and

BCR of diameter d1 and d2 respectively are inscribed in the semicircle

PQR such that the sum of d1 and d2 is equal to 10 cm.

Complete

Table 1 by using various values of d1 and the corresponding values of

d2. Hence, determine the relation between the lengths of arcs PQR, PAB

and BCR.Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and

BCR of diameter d1 and d2 respectively are inscribed in the semicircle

PQR such that the sum of d1 and d2 is equal to 10 cm.

Complete

Table 1 by using various values of d1 and the corresponding values of

d2. Hence, determine the relation between the lengths of arcs PQR, PAB

and BCR.

The length of arc (s) of a circle can be found by using the formula

where r is the radius.

From the table, we can conclude that

Length of arc PQR = Length of arc PAB + Length of arc BCR

*(b)*

Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB,

BCD and DER of diameter d1, d2 and d3 respectively are inscribed in the

semicircle PQR such that the sum of d1, d2 and d3 is equal to 10 cm.

(i)

Using various values of d1 and d2 and the corresponding values of d3,

determine the relation between the lengths of arcs PQR, PAB, BCD and

DER. Tabulate your findings.Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB,

BCD and DER of diameter d1, d2 and d3 respectively are inscribed in the

semicircle PQR such that the sum of d1, d2 and d3 is equal to 10 cm.

(i)

Using various values of d1 and d2 and the corresponding values of d3,

determine the relation between the lengths of arcs PQR, PAB, BCD and

DER. Tabulate your findings.

Again, we use the same formula to find the length of arc of PQR, PAB, BCD and DER.

This time, the table is a big one. You can click on the table to change to the full size view.

Again, we can conclude that:

Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of arc CDR

*(ii)*

Based on your findings in (a) and (b), make generalisations about the

length of the arc of the outer semicircle and the lengths of arcs of

the inner semicircles for n inner semicircles where n = 2, 3, 4....Based on your findings in (a) and (b), make generalisations about the

length of the arc of the outer semicircle and the lengths of arcs of

the inner semicircles for n inner semicircles where n = 2, 3, 4....

Base on the findings in the table in(a) and (b) above, we conclude that:

The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles.

*(c) For different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.*

Diagram above shows a big semicircle with n number of small inner circle. From the diagram, we can see that

The length of arc of the outer semicircle

The sum of the length of arcs of the inner semicircles

Factorise π/2

Substitute

We get,

where d is any positive real number.

We can see that

As a result, we can conclude that

The

length of the arc of the outer semicircle is equal to the sum of the

length of arcs of any number of the inner semicircles. This is true for

any value of the diameter of the semicircle.

In

other words, for different values of diameters of the outer semicircle,

show that the generalisations stated in b (ii) is still true.

*To be CONTINUE ...................***Giraffe**- 校长
- 帖子数 : 542

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来自 : johor

注册日期 : 09-07-22

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## 回复： Form 5 Add Math Project 2009

(c)

y = -πx^2/4 + 5πx/2

Change it to linear form of Y = mX + C.

y/x = -πx/4 + 5π/2

Y = y/x

X = x

m = -π/4

C = 5π/2

Thus, plot a graph of y/x against x and draw the line of best fit.

Find the value of y/x when x = 4.5 m.

Then multiply y/x you get with 4.5 to get the actual value of y.

3 (d) We need to get the largest value of y so that the cost of constructing the garden is minimum.

y = -πx^2/4 + 5πx/2

dy/dx = -πx/2 + 5π/2

(d^2)y/dx^2 = -π/2 <--- y has a maximum value.

At maximum point, (d^2)y/dx^2 = 0.

-πx/2 + 5π/2 = 0

πx/2 = 5π/2

x = 5 m

x = 5 m:

maximum value of y = -π(5^2)/4 + 5π(5)/2

= 6.25π m^2

y = -πx^2/4 + 5πx/2

= -π/4(x^2 - 10x)

= -π/4(x^2 - 10x + 25 - 25)

= -π/4[(x - 5)^2 - 25]

= -π/4(x - 5)^2 + 25π/4

y is a n shape graph as a = -π/4.

Hence, it has a maximum value.

When x = 5 m, maximum value of the graph = 6.25π m^2.

3 (e)

The keywords are:

The

principal suggested an additional of 12 semicircular flower beds to the

design submitted by the Mathematics Society. (n = 12)

The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m)

The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)

The diameter of the flower beds are increased by a constant value successively. (d = ?)

S12 = (n/2)[2a + (n - 1)d]

10 = (12/2)[2(0.3) + (12-1)d]

= 6(0.6 + 11d)

= 3.6 + 66d

66d = 6.4

d = 16/165

Since the first flower bed is 0.3 m,

Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:

131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m

If there is any mistake please kindly PM or e-mail me.

I haven't done my project work yet but this is my brief working on it.

Let me know my mistake so that I won't get wrong when I started doing it.

Part 2 (b) (ii):

You need to make a generalization by induction for findings in (a) and (b) (ii) but not a statement.

I suggest,

(s out) = ∑ n (s in), n = 2, 3, 4, ......

where,

s in = length of arc of inner semicircle

s out = length of arc of outer semicircle

Part 2 (c):

You need to take at least 2 different values of diameters for the outer semicircle.

You need to show your tables for each value of those diameters.

This part is only to prove that your generalization stated in Part 2 (b) (ii) is still true.

But it is good to make a proof for it.

__Linear Law__y = -πx^2/4 + 5πx/2

Change it to linear form of Y = mX + C.

y/x = -πx/4 + 5π/2

Y = y/x

X = x

m = -π/4

C = 5π/2

Thus, plot a graph of y/x against x and draw the line of best fit.

Find the value of y/x when x = 4.5 m.

Then multiply y/x you get with 4.5 to get the actual value of y.

3 (d) We need to get the largest value of y so that the cost of constructing the garden is minimum.

__Method 1: Differentiation__y = -πx^2/4 + 5πx/2

dy/dx = -πx/2 + 5π/2

(d^2)y/dx^2 = -π/2 <--- y has a maximum value.

At maximum point, (d^2)y/dx^2 = 0.

-πx/2 + 5π/2 = 0

πx/2 = 5π/2

x = 5 m

x = 5 m:

maximum value of y = -π(5^2)/4 + 5π(5)/2

= 6.25π m^2

__Method 2: Completing the Square__y = -πx^2/4 + 5πx/2

= -π/4(x^2 - 10x)

= -π/4(x^2 - 10x + 25 - 25)

= -π/4[(x - 5)^2 - 25]

= -π/4(x - 5)^2 + 25π/4

y is a n shape graph as a = -π/4.

Hence, it has a maximum value.

When x = 5 m, maximum value of the graph = 6.25π m^2.

3 (e)

__Arithmetic Progression__The keywords are:

The

principal suggested an additional of 12 semicircular flower beds to the

design submitted by the Mathematics Society. (n = 12)

The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m)

The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)

The diameter of the flower beds are increased by a constant value successively. (d = ?)

S12 = (n/2)[2a + (n - 1)d]

10 = (12/2)[2(0.3) + (12-1)d]

= 6(0.6 + 11d)

= 3.6 + 66d

66d = 6.4

d = 16/165

Since the first flower bed is 0.3 m,

Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:

131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m

If there is any mistake please kindly PM or e-mail me.

I haven't done my project work yet but this is my brief working on it.

Let me know my mistake so that I won't get wrong when I started doing it.

**To post 6:**(This is what my teacher said.)Part 2 (b) (ii):

You need to make a generalization by induction for findings in (a) and (b) (ii) but not a statement.

I suggest,

(s out) = ∑ n (s in), n = 2, 3, 4, ......

where,

s in = length of arc of inner semicircle

s out = length of arc of outer semicircle

Part 2 (c):

You need to take at least 2 different values of diameters for the outer semicircle.

You need to show your tables for each value of those diameters.

This part is only to prove that your generalization stated in Part 2 (b) (ii) is still true.

But it is good to make a proof for it.

**Giraffe**- 校长
- 帖子数 : 542

1018

来自 : johor

注册日期 : 09-07-22

年龄 : 26

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