Additional Mathematics Project

PROJECT WORK FOR ADDITIONAL MATHERMATICS 2009

CI RCLE IN OU R DA ILY LIFE

1

INDEX B IL.

CONTEN TS

1

PART

2

PAGES

1

•

INTRODUCTIONS

3

•

INFO

6

PART 2 METHODS

3

•

QUESTIONS 2(a)

7

•

QUESTIONS 2(b)

8

PART 3 •

CONJECTURES

10

•

CONCLUSION

13

2

IN TRODU CTIONS There are a lot of things around us related to circles or parts of a circles. We need to play with circles in order to complete some of the problems involving circles. In this project I will use the principles of circle to design a garden to beautify the school.

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De fin ition In Euclidean plane geometry, π is defined as the ratio of a circle's circumference to its diameter:

The ratio C/d is constant, regardless of a circle's size. For example, if a circle has twice the diameter d of another circle it will also have twice the circumference C, preserving the ratio C/d. Area of the circle = π × area of the shaded square Alternatively π can be also defined as the ratio of a circle's area (A) to the area of a square whose side is equal to the radius

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History The ancient Babylonians calculated the area of a circle by taking 3 times the square of its radius, which gave a value of pi = 3. One Babylonian tablet (ca. 1900–1680 BC) indicates a value of 3.125 for pi, which is a closer approximation. In the Egyptian Rhind Papyrus (ca.1650 BC), there is evidence that the Egyptians calculated the area of a circle by a formula that gave the approximate value of 3.1605 for pi. The ancient cultures mentioned above found their approximations by measurement. The first calculation of pi was done by Archimedes of Syracuse (287–212 BC), one of the greatest mathematicians of the ancient world. Archimedes approximated the area of a circle by using the Pythagorean Theorem to find the areas of two regular polygons: the polygon inscribed within the circle and the polygon within which the circle was circumscribed. Since the actual area of the circle lies between the areas of the inscribed and circumscribed polygons, the areas of the polygons gave upper and lower bounds for the area of the circle. Archimedes knew that he had not found the value of pi but only an approximation within those limits. In this way, Archimedes showed that pi is between 3 1/7 and 3 10/71. A similar approach was used by Zu Chongzhi (429–501), a brilliant Chinese mathematician and astronomer. Zu Chongzhi would not have been familiar with Archimedes’ method—but because his book has been lost, little is known of his work. He calculated the value of the ratio of the circumference of a circle to its diameter to be 355/113. To compute this accuracy for pi, he must have started with an inscribed regular 24,576-gon and performed lengthy calculations involving hundreds of square roots carried out to 9 decimal places.

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IN FO Part

2 (a)

Q

C A

P d1

R

B

d2 10 cm

Diagram 1 shows a semicircle PQR of diameter 10cm. Semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in PQR such that the sum of d1 and d2 is equal to 10cm. By using various values of d1 and corresponding values of d2, I determine the relation between length of arc PQR, PAB, and BCR. Using formula: Arc of semicircle = ½πd

d1 (cm) 1 2 3 4 5

d2 (cm) 9 8 7 6 5

Length of arc PQR in terms of π (cm) 5 π 5 π 5 π 5 π 5 π

Length of arc PAB in terms of π (cm) ½ π π 3/2 π 2 π 5/2π

Length of arc BCR in terms of π (cm) 9/2 π 4 π 7/2 π 3 π 5/2 π

From the Table 1 we know that the length of arc PQR is not affected by the different in d1 and d2 in PAB and BCR respectively. The relation between the length of arcs PQR , PAB and BCR

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is that the length of arc PQR is equal to the sum of the length of arcs PAB and BCR, which is we can get the equation:

Let d1= 3, and d2 =7

5π 5π

SPQR

=

SPAB

+

SBCR

SPQR

=

SPAB

+

SBCR

5π = ½ π(3) + ½ π(7) 5π = 3/2 π + 7/2 π = 10/2 π = 5 π

Q

E C

A 2(B)ii

P d1

B

D

d2 10 7

R d3

d1 1 2 2 2 2

SPQR

d2 2 2 3 4 5

=

d3 7 6 5 4 3

SPAB +

SPQR 5π 5π 5π 5π 5π

SBCD +

5 π

= =

π

+

SBCD π π 3/2 π 2 π 5/2 π

SDER 7/2 π 3 π 5/2 π 2 π 3/2 π

SDER

Let d1 = 2, d2 = 5, d3 = 3 5 π

SPAB 1/2 π π π π π

SPQR = 5/2 π

+

SPAB

+

SBCD

+

SDER

3/2 π

5 π

B (ii) The length of arc of outer semicircle is equal to the sum of the length of arc of inner semicircle for n = 1,2,3,4,…. Souter = S1 + S2 + S3 + S4 + S5

CONCLUSIONS Base on the findings in the table in (a) and (b) above, we conclude that: The length of the arc of the outer semicircle = the sum of the length of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4… Or (s out) = Σ n (s in), n = 2, 3, 4,...... where, s in = length of arc of inner semicircle s out = length of arc of outer semicircle (C) Assume that the diameter of the outer semicircle is 30cm and 4 semicircles are inscribed in the outer semicircle such that the sum of d1 (arc1), d2 (arc2), d3 (arc3), d4 (arc4) is equal to 30cm. Again, by using various values of d1, d2, d3, and the corresponding values of

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d4, in which d1 + d2 + d3 + d4 = 30cm, I had make a table to determine the relation between the lengths of arc1, arc2, arc3, and arc4. All the results are tabulated in Table 3 below: d1 (cm)

d2 (cm)

d3 (cm)

d4 (cm)

10 12 14 15

8 3 8 5

6 5 4 3

6 10 4 7

Length of arc of outer semicircl e in terms of π (cm) 15 π 15 π 15 π 15 π

Length of arc 1 in terms of π (cm)

Length of arc 2 in terms of π (cm)

Length of arc 3 in terms of π (cm)

Length of arc 4 in terms of π (cm)

5π 6π 7π 15/2 π

4π 3/2 π 4π 5/2 π

3π 5/2 π 2π 3/2 π

3π 5π 2π 7/2 π

Table 3

From the Table 3, we know that the length of arc PQR is not affected by the different in d1, d2, d3, and d4 in arc1, arc2, arc3, and arc4 respectively. The relation between the length of arc of outer semicircle, arc1, arc2, arc3, and arc4 is that the length of arc of outer semicircle is equal to the sum of the length of arc1, arc2, arc3, and arc4 which the equation is: S ou ter se mi circl e = S ar c1 + S ar c2 + S ar c3 + S ar c4 Let d1=10, d2=8, d3=6, d4=6 Souter semicircle = Sarc1 + Sarc2 + Sarc3 + Sarc4 15 π = 5 π + 4 π + 3 π + 3 π 15 π = 15 π Therefore, we can conclude that length of arc of outer semicircle is equal to the sum of the length of arc of inner semicircle for x = 2, 3, 4, 5…..

S ou ter

=

S 1 + S 2 + S 3 + S 4 + S 5 .. ... .

From the table, we can conclude that Length of arc PQR = Length of arc PAB + Length of arc BCR CONJE CTUR ES 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 flower and the two inner semicircles are fish ponds.

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(a)

The area of the flower plot is y m2 and the diameter of one of the fish pond is x cm.

Area of flower plot = y m2 y y y y y y y y

(25/2) π - (1/2(x/2)2 π + 1/2((10-x )/2)2 π) (25/2) π - (1/2(x/2)2 π + 1/2((100-20x+x2)/4) π) (25/2) π - (x2/8 π + ((100 - 20x + x2)/8) π) (25/2) π - (x2π + 100π – 20x π + x2π )/8 (25/2) π - ( 2x2 – 20x + 100)/8) π (25/2) π (( x2 – 10x + 50)/4) (25/2 (x2 - 10x + 50)/4) π ((10x – x2)/4) π

= = = = = = = =

Therefore, the area of flower plot is equal to ((10x – x2)/4) π. (b) By using π = 22/7, we can find the diameters of the two fish ponds if the area of the flower plot is 16.5 m2. Area of flower plot

= =

16.5 m2 ((10x – x2)/4) π

16.5 = ((10x – x2)/4) π 66 = (10x - x2) 22/7 66(7/22) = 10x – x2 0 = x2 - 10x + 21 0 = (x-7) (x – 3) x = 7 , x = 3 Therefore, the diameter of fish pond E is 3 m while the diameter of fish pond F is 7 m. (c)

We can reduce the non-linear equation obtained in (a) to simpler linear form. ((10x – x2)/4) π ((10x – x2)/4x) π (10/4 - x/4) π

y = y/x = y/x =

Y/x

8.0

To determine the area of the flower plot, we have to plot a straight line graph by using the 7.0 equation: 6.0

y/x =

(10/4 - x/4) π 5.0

x

1

2

3

4

5

6

7

10

4.0

3.0

2.0 0

1

2

3

4

5

6

7

X

y/ x

7.1

6.3

5.5

4.7

3.9

3.1

2.4

From the graph, we can determine the area of the flower pot if the diameter of one of the fish pond is 4.5 m. Area of flower plot = y When x = 4.5, y/x = 4.3 y = y/x * x y = 4.3 * 4.5 y = 19.35m2 Therefore, the area of the flower pot if the diameter of one of the fish pond is 4.5 m is equal to 19.35m2. (d) The cost of constructing the fish ponds is higher than that of the flower plot. There are two methods to determine the area of the flower plot such that the cost of constructing the garden is minimum. The area of the flower plot can be determined by using differentiation method and completing square method. (i)

Differentiation method: dy/dx

=

((10x-x2)/4) π = ( 10/4 – 2x/4) π = 5/2 π – x/2 π

Since the cost of constructing the garden is minimum, thus dy/dx is equal to 0. 0 = 5/2 π – x/2 π 5/2 π = x/2 π x = 5

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(ii)

Completing Square method: y = = = = =

((10x – x2)/4) π 10x/4 π - x2/4 π -1/4 π (x2 – 10x) -1/4 π (x – 5)2 - 52 -1/4 π (x - 5)2 – 25

Since the cost of constructing the garden is minimum, thus (x-5)2 is equal to 0. x – 5 = 0 x = 5 Therefore, the area of the flower pot is 5 m2.

(e) The principle 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. We can determine the diameter of the remaining flower beds by using the formula :

S n = n/ 2 ( 2a + (n – 1) d n = 12, a = 30cm, S12 = 1000cm 1000 = 12/2 (2(30) + (12 – 1) d) 1000 = 6 (60 + 11d) 1000 = 360 + 66d 1000 – 360 = 66d 640 = 66d d = 9.697

Tn (flower bed) T1 T2 T3 T4 T5 T6 T7 T8 T9 T10 T11 T12

Diameter (cm) 30 39.697 49.394 59.091 68.788 78.485 88.182 97.879 107.576 117.273 126.97 136.667

Since d is equal to 9.697, thus, we can find the diameter of the following flower beds The diameter of the remaining flower beds = T 5 + T6 + T7 + T8 + T9 + T10 + T11 +T12 = 68.788 + 78.485 + 88.182 + 97.879 + 107.576 +

12

117.273 + 126.97 + 136.667 =821.82 cm Therefore, the diameter of the remaining flower beds is equal to 821.82 cm.

CON CL US IO N 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.

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