Add Maths Question

PART 2 Relation between the lengths of arcs PQR, PAB and BCR a) Diagram 1 shows a semicircle PQR of diameter 10cm.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

d1(cm)

d2(cm)

1.0 1.2 1.4 1.6 1.8 2.0

9.0 8.8 8.6 8.4 8.2 8.0

L.O.A PQR in terms of п(cm) 5п 5п 5п 5п 5п 5п

L.O.A PAB in terms of п(cm) 0.5п 0.6п 0.7п 0.8п 0.9п 1.0п

L.O.A BCR in terms of п(cm) 4.5п 4.4п 4.3п 4.2п 4.1п 4.0п

2.2 2.4 2.6 2.8

7.8 7.6 7.4 7.2

5п 5п 5п 5п

1.1п 1.2п 1.3п 1.4п

3.9п 3.8п 3.7п 3.6п

TABLE 1 b) Diagram 2 shows a semicircle PQR of diameter 10cm.Semicircle 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 10cm.

DIAGRAM 2

i) Relation between the lengths of arcs PQR, PAB, BCD and DER:

d1(cm)

d2(cm)

d3(cm)

2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1

3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 4.6 4.8

5.0 4.9 4.8 4.7 4.6 4.5 4.4 4.3 4.2 4.1

ii)

L.O.A PQR in terms of п(cm) 5п 5п 5п 5п 5п 5п 5п 5п 5п 5п

L.O.A PAB in terms of п(cm) 1п 19/20п 9/10п 17/20п 4/5п 3/4п 7/10п 13/20п 3/5п 11/20п

L.O.A BCD in terms of п(cm) 3/2п 8/5п 17/10п 9/5п 19/10п 2п 21/10п 11/5п 23/10п 12/5п

L.O.A DER in terms of п(cm) 5/2п 49/20п 12/5п 47/20п 23/10п 9/4п 11/5п 43/20п 21/10п 41/20п

L.O.A 1 + L.O.A 2 = L.O.A 3

Generalizations: The total L.O.A of the inner semicircles inscribed inside the outer semicircles is equal to the L.O.A of the outer semicircles.

c) Different values of diameters of the outer semicircle: d1(cm) d2(cm) L.O.A PQR in L.O.A PAB in terms of terms of п(cm) п(cm) 2.0 12.0 7п 1п 2.2 11.8 7п 11/10п 2.4 11.6 7п 6/5п 2.6 11.4 7п 13/10п 2.8 11.2 7п 7/5п 3.0 11.0 7п 3/2п 3.2 10.8 7п 8/5п 3.4 10.6 7п 17/10п 3.6 10.4 7п 9/5п

L.O.A BCR in terms of п(cm) 6п 59/10п 29/5п 57/10п 28/5п 11/2п 27/5п 53/10п 26/5п

3.8

10.2

7п

19/10п

51/10п

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 ) Area of garden = Πr² 2 =Π(5)² 2 =25Π 2 y = 25Π – 1 -1 2 x²Π 2(10-x)²Π 2 2

y = 25Π – Π(x)² - Π(10 – x)² 2 2 2 2 2 = 25Π – Π (x ² ) – Π (100Π - 20Πx - x² 2 4 4 2 2 = 25Π- Π x² - 100Π - 20Πx - x² 2 8 8 y = 23Π - Πx² - 100Π + 20Πx - x² 2 8 8 8 8 y = (100-x²- 100 + 20Π- x² ) 8 y = (-2x² + 20x ) 8 Π = ( -x² + 5x ) 4 2 y = -Π x² + 5Πx 4 2

b ) 16.5 = Π (-2x² + 20x ) 8 16.5 = Π( -2x² +20x ) 8 16.5 = -2x² + 20x Π 8 16.5 = -2x² + 20x (22 ) 8 7 115.5 = -2x² + 20x 22 8 5.25 = -2x² + 20x 8 42 = -2x²-20x 21 = -x²- 10x x² + 10x + 21 = 0 ( x-7)

(x-3 )

x-7 = 0 x= 7

x-3 = 0 ,

x= 3

c) X Y Y X

1 7.068 7.068

y = 5Πx - Πx² 2 4 Y = 5Π – Πx X 2 4 Y = -Π x + 5 Π X 4 2 Y = Mx + C Y=Y X m=-∏ 4 X=x C = 5п 2

2 6.283 3.141

3 5.497 1.832

4 4.712 1.178

5 3.926 0.785

6 2.356 0.523

7 1.570 0.336

GRAPH OF Y AGAINST X 8

7

6

5

y 4 1

3

2

1

0 1

2

3

4

5

The value of y when x is 4.5 = 4

6

d ) y = 5пx - пx² 2 4 Method 1 : Diffrentiation dy = 5п – п x² = 0 dx 2 2 5п = пx 2 2 X=5 Therefore, max area for y when x = 5 Y maximum = 5п(5) – п(5)² 2 2 =19.634m² Method 2 : Completing the square y = -пx² - 5пx 4 2

,

÷п 4

= -п( x² - 10x ) 4 = -п(x² + (-10 )² - (-10 )² 4 2 2 = -п( x – 5 ²) - 5²п 4 4 -п( x – 5 )² - 25п 4 4 Therefore,max area for y whwn x = 5 Y maximium = 25п 4 =19.634m