JURNAL MATEMATIK TAMBAHAN

dius, in m, for the following six

Answer for Question 2

Radius in m for :
1. Eye on Malaysia = 26m
2. Wiener Riesenrad, Vienna = 30.4
3. The London Eye = 60m
4. Tianjin Eye, China = 55m
5. High Roller, Las Vegas = 79.25m
6. The Singapore Flyer = 75m

48m

Question 3

If the coordinates of the centre
determine

i. the circumference of each Fe

ii. the area, in m2 , covered by e
complete oscillation,

iii. the equation for each Ferris

of each Ferris wheel is (0, 0),

erris wheel, in m,
each Ferris wheel in one

wheel

i(.C)itrhcuemcifrecruemncfeer=e2n c e)of each Ferr
1. Eye on Malaysia

=2(3.142)(26)
=163.38m

2. Wiener Riesenrad, Vienna
=2(3.142)(30.48)
=191.54m

3. The London Eye
=2(3.142)(60)
=377.04m

ris wheel, in m,

4. Tianjin, China
=2(3.142)(55)
=345.62m

5. High Roller, Las Vegas
=2(3.142)(79.25)
=498.01m

6. The Singapore Flyer
=2(3.142)(75)
=471.3m



ioi.s)ctilhlaetaiorne,aA, irneamo2f,ccirocvlee,re d=by e a ch
1. Eye on Malaysia
=(3.142)(26)2
=2123.99m2

2. Wienner Riesenrad, Vienna
=(3.142)(30.48)2
=2919.01m2

3. The London Eye
=(3.142)(60)2
=11311.2m2

h Ferris wheel in one complete

3. Tianjin Eye, China
=(3.142)(55)2
=9504.55m2

4. High Roller, Las Viegas
=(3.142)(79.25)2
=19733.53m2

5. The Singapore Flyer
=(3.142)(75)2
=17673.75m2



iii.) the equation for each

1. Eye on Malaysia
Using point (0,0)

(x − 0)2 + (y − 0)2 = 26
x2+y2=262
x2+y2=676

2. Wienner Riesenrad, Vienna
Using point (0,0)

(x − 0)2 + (y − 0)2 =30.48
x2+y2=30.482
x2+y2= 929.03

h Ferris wheel

3. The London Eye
Using point (0,0)

(x − 0)2 + (y − 0)2 =60
x2+y2=602
x2+y2=3600

4. Tianjin Eye, China

Using point (0,0)

(x − 0)2 + (y − 0)2 =55
x2+y2=552
x2+y2=3025



5. High Roller, Las Vegas
Using point (0,0)

(x − 0)2 + (y − 0)2 = 79.25
x2+y2=79.252
x2+y2=6280.56

6. The Singapore Flyer
Using point (0,0)

(x − 0)2 + (y − 0)2 = 75
x2+y2=752
x2+y2=5625