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

where r is the radius. The result is as below:

table-1.png (31.58 KB) Viewed 10912 times 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. 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.

table-2.png (52.46 KB) Viewed 10949 times 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.... 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.

circle6.png (10.23 KB) Viewed 10850 times

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.

Since post 6 had answered Part 1 and Part 2, I will just show the solution for Part 3. As in post 2 had answered question 3 (a) and 3 (b), I will briefly explain question 3 (c), 3 (d) and 3 (e). Part 3 is a little bit mind twisting. 3 (c) 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=5m 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.

[quote="cschuah92"]Since post 6 had answered Part 1 and Part 2, I will just show the solution for Part 3.

As in post 2 had answered question 3 (a) and 3 (b), I will briefly explain question 3 (c), 3 (d) and 3 (e). Part 3 is a little bit mind twisting. 3 (c) 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 (e) Arithmetic Progression a = 30 cm = 0.3 m S12 = 10 m n = 12 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 Sequence of arithmetic progression: 3/10, 131/330, 163/330, 13/22, 227/330, 259/330, 97/110, 323/330, 71/66, 129/110, 419/330, 41/30