Mathematics · May / June 2025 · Paper 11 · Question 6
YMA01
As printed in the paper
The question as text
6. A Not to scale O C B Figure 2 The shaded area in Figure 2 shows the plan view of a helicopter landing pad. The area consists of the major sector AOB of a circle centre O joined to a triangle AOC. Given that • AO = OB = 15 m • BC = 2 m • CBO is a straight line • angle ACO = 0.6 radians
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
17
15
Angle COA =
π
−
0.6 ''0.6944'' 1.847
−
=
*
1
1
(
) or just 1.847
6(b)[3]
2
2
Attempts
r θ
=
×
15
×θ where
θ
=
2
π
−
1.847
OR
2
2
1
1
(
)
attempts
sin
=
15 17sin 1.847
ab
C
×
×
2
2
1
1
(
)(
)
2
2
Attempts
r θ
=
×
15
×
2
π−
1.847
≈
499
AND
2
2
1
1
(
)(
)
ab
sin
C
=
×
15 17sin 1.847
×
≈
122.7
AND adds
2
2
(awrt) 622 m2
(
) or just 1.847
×θ where
θ
=
2
π
−
1.847
OR
rθ=
15
6(c)[3]
(
)
2
2
2
(
)(
)
AC =
15
+
17
−×
2 15 17cos 1.847
×
≈
653
(
)(
)
Attempts
rθ=
15
×
2
π−
1.847
=
66.54
AND
2
2
(
)
AC =
15
+
17
−×
2 15 17cos 1.847
×
92.1 + 2 = (awrt) 94.1 m
Note: Work must be completed in the appropriate parts to gain credit, but if no labelling is given, apply
benefit of doubt that area work is for (b) and perimeter work is for (c).
(a)
M1: Writes down a correct sine rule and attempts to find angle CAO. Alternatively, writes down a correct
cosine rule and attempts to find AC :
34cos0.6
(
34cos0.6
)
2
4
64
±
−
×
1
5
2
=
1
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