Project Work For Add Math

Question A We use the formula to find the length of arc of PQR, PAB and BCR

d1(cm)

1 2 3 4 5 6 7 8 9

d2(cm)

9 8 7 6 5 4 3 2 1

Length of arc

Length of arc

Length of arc

PQR in terms

PAB in terms

BCR in terms

of π (cm) 5π 5π 5π 5π 5π 5π 5π 5π 5π

of π (cm) 0.5π 1.0π 1.5π 2.0π 2.5π 3.0π 3.5π 4.0π 4.5π

of π (cm) 4.5π 4.0π 3.5π 3.0π 2.5π 2.0π 1.5π 1.0π 0.5π

From the table, we can conclude that Length of arc PQR = Length of arc PAB + Length of arc BCR

Question B (i) We use the same formula to find the length of arc of PQR, PAB, BCD and DER

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d1(cm)

1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8

d2(cm)

1 2 1 3 2 1 4 3 2 1 5 4 3 2 1 6 5 4 3 2 1 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1

d3(cm)

8 7 7 6 6 6 5 5 5 5 4 4 4 4 4 3 3 3 3 3 3 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1

Length of

Length of

Length of

Length of

arc PQR in

arc PAB in

arc BCD in

arc DER in

terms of π

terms of π

terms of π

terms of π

(cm) 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π 5π

(cm) 0.5π 0.5π 1.0π 0.5π 1.0π 1.5π 0.5π 1.0π 1.5π 2.0π 0.5π 1.0π 1.5π 2.0π 2.5π 0.5π 1.0π 1.5π 2.0π 2.5π 3.0π 1.0π 1.0π 1.5π 2.0π 2.5π 3.0π 3.5π 0.5π 1.0π 1.5π 2.0π 2.5π 3.0π 3.5π 4.0π

(cm) 0.5π 1.0π 0.5π 1.5π 1.0π 0.5π 2.0π 1.5π 1.0π 0.5π 2.5π 2.0π 1.5π 1.0π 0.5π 3.0π 2.5π 2.0π 1.5π 1.0π 0.5π 3.5π 3.0π 2.5π 2.0π 1.5π 1.0π 0.5π 4.0π 3.5π 3.0π 2.5π 2.0π 1.5π 1.0π 0.5π

(cm) 4.0π 3.5π 3.5π 3.0π 3.0π 3.0π 2.5π 2.5π 2.5π 2.5π 2.0π 2.0π 2.0π 2.0π 2.0π 1.5π 1.5π 1.5π 1.5π 1.5π 1.5π 1.0π 1.0π 1.0π 1.0π 1.0π 1.0π 1.0π 0.5π 0.5π 0.5π 0.5π 0.5π 0.5π 0.5π 0.5π

Again we can conclude that 6

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

Question B (ii) Based 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.

Question C 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 7

We can see that (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 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.

Question A

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Question B 9

Question C 10

By Linear Law y = -πx2/4 + 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 x

2

4

6

8

10

y

4

6

6

4

0

x

2

4

6

8

10

y x

2

1.5

1

0.5

0

x-axis : 2cm to 2 unit y-axis : 2cm to 0.5 unit

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