Challenging P6 2010 HHK P2 Q12

Fig 1 Area of fig: 6

Fig 2 Area of fig: 10
Fig 3 Area of fig: 16
Fig 4 Area of fig: 24

(c) Find the area of figure 100.
Ans: 10104.
For Patterns, try to make use of the figures given to look for specific patterns.
F1 --> 6 = 2 + 4
= 1 + 1 + 4
= 1 + (1 x 1) + 4

F2 --> 10 = 6 + 4
= 2 + 4 + 4
= 2 + (2 x 2) + 4

F3 --> 16 = 12 + 4
= 3 + 9 + 4
= 3 + (3 x 3) + 4

F4 --> 24 = 20 + 4
= 4 + 16 + 4
= 4 + (4 x 4) + 4

Fn --> n + (n x n) + 4

Circle P6 2010 SA2 HHK P2 Q13

The figure below shows 2 quarter circles and a rectangle. The radius of the big quarter circle is 8 cm. The radius of the small quarter circle is 4 cm. Find the difference in area between the two shaded parts X and Y. Use the calculator value of pi and give your answer correct to 1 decimal place.

1/4 x ∏ x 8 x 8 = 16 pi

1/4 x ∏ x 4 x 4 = 4 pi

16 pi - 4 pi =  12 pi  B + X

8 x 4 = 32       B + Y

12 pi  - 32 = 5.7 cm2

It was 5.7 cm2.

Please calculate ?? as I do not have a proper calculator.

Average P6 2010 SA2 HHK P2 Q17

The average length of 6 ropes was 80 cm. The average length of 4 of the ropes A,B, C and D was 15 cm more than the average length of the remaining 2 ropes E and F. (a) Find the average length of ropes E and F. Give your answer in metres. (b) If Rope E had been 18 cm shorter and Rope F had been 12 cm longer, they would have been of the same length. Find the actual length of Rope E. [5 marks][Answer given: 0.7m, 85 cm]


80 x 6 = 480

15 x 4 = 60

480 - 60 = 420

420 / 6 = 70 cm
            = 0.7 m

a) It was 0.7 m.


E [         ][    ][   18   ]      ] 70 x 2
F [         ][12]                   ] = 140 cm

2 u --> 140 - 12 - 18 = 110
1 u --> 110/2 = 55

55 + 12 + 18 = 85 cm

b) It was 85 cm.