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Mathematics · May / June 2022 · Paper 11 · Question 8

YMA01

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The question as text

8. C D B A O Figure 3 Figure 3 shows a sketch of the outline of the face of a ceiling fan viewed from below. The fan consists of three identical sections congruent to OABCDO, shown in Figure 3, where • OABO is a sector of a circle with centre O and radius 9 cm • OBCDO is a sector of a circle with centre O and radius 84 cm π radians angle AOD = 2 • 3 Given that the length of the arc AB is 15 cm,

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  • 8(a)[3]

    and being the angle subtended by arc CD

    5

    (

    )

    1

    5

    =

    9×

    θ

    θ

    =

    =

    1.6

    7

    3

    2

    π

    5

    (

    )

    Therefore

    =

    −

    =

    0.4277...

    3

    3

    2

    π

    5

    8

    4

    35.929...

    35.9cm

    So length of arc CD =

    ×

    −

    =

    =

    (1 d.p.)* CSO

    3

    3

    Perimeter

    =

    3×

    (15 35.9...)

    +

    + ×

    6 (84 9)

    −

  • 8(b)[2]

    = awrt 603 cm (602.787…)

    1

    "

    2

    π−

    5

    "

  • 8(c)[5]

    84

    2

    = awrt (1510)

    FOR EXAMPLE Area of a “blade” is

    ×

    ×

    2

    3

    1

    5

    "

    "

    Area of sector of inner circle between “blades” is

    ×

    9

    2

    ×

    = (67.5)

    2

    3

    1

    "

    2

    π

    5

    "

    1

    "

    5

    "

    Total area is

    3

    ×

    84

    2

    ×

    −

    +

    ×

    9

    2

    ×

    =

    ...(4729.577764cm

    2

    )

    2

    3

    3

    2

    3

    So area is awrt 0.473 m2 or awrt 4730 cm2

    Notes:

    (a)

    M1: Correct use of the arc length formula to find the angle subtended by arc AB.

    15

    9

    ..

    .

    Attempts

    =

    ×

    θ

    θ

    =

    Don't be concerned by what the angle is called

    dM1: Correct method to find the angle subtended by arc CD using their angle for arc AB.

    1

    5

    Note that

    =

    2

    π

    −×

    3

    is also correct. It is dependent upon the previous M

    3

    3

    A1*: CSO Arrives at 35.9 with a correct value to at least 2 d.p. (rounded or truncated) seen first.

    2

    π

    5

    (

    )

    Alternatively sight of

    8

    4

    ×

    −

    or

    84×

    awrt 0.4277

    followed by

    35.9 cm is fine

    3

    3

    2

    π

    5

    or 2

    π

    Note that there are equivalent methods such as

    84

    ×

    −

    84

    ×

    =

    35.9

    ×

    84

    −

    140

    =

    3

    3

    3

    (b)

    M1: Correct method to find the perimeter, it should include all six arcs and radial edges.

    Look for

    3×

    15

    + ×

    3 35.9

    + ×

    6 ...

    If no method is seen it is implied by awrt 603

    A1: For awrt 603 (cm). The units need not be given.

    (c) This part is now being marked M1 dM1 A1 ddM1 A1

    Please look through all of the solution first. The marks can be awarded in the following way.

    M1: A correct attempt at any relevant area

    dM1: A correct attempt at a corresponding area that can be combined with the first area in some way

    to find the area of the fan. FT on angles found in part (a). Dependent upon previous mark

    A1: Both areas correct. They do not need to be calculated but the angles must be correct to 3sf

    ddM1: A correct combination of areas to find the area of the fan

    A1: awrt 0.473 m2 or awrt 4730 cm2. Must include the units. ISW after a correct answer

    (i) 2p

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