Mathematics · Oct / Nov 2022 · Paper 11 · Question 8

YMA01

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The question as text

8. Diagram NOT to scale X 8 m B C π rad 3 m 6 O π rad 2 3 3 m A Figure 2 Figure 2 shows the plan view of a design for a pond. The design consists of a sector AOBX of a circle centre O joined to a quadrilateral AOBC. • BC = 8 m • OA = OB = 3 m angle AOB is 2 π radians • 3 π radians • angle BCA is 6

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8(a)[5]
2
π
4
π
2
π−
=
3
3
1
4
Area of sector
=
×
3
2
×
"
3π
"
=
6
π
(m2)
2
4
Length of arc
= ×
3 "
3π
"
Perimeter
=
4
π
+ (m)
6
(ii)
1
2
9 3
×
3
2
×
sin
π
=
(m2)
2
3
4
2
Eg
2
=
3
2
+
3
2
−×××
2 3 3 cos
π
2
=
27
=
3 3
(m) *
AB
AB
AB
3
π
sin
sin
BAC
6
=
sin
BAC
=
.... or
BAC
=
...
8
3 3
4 3
sin
BAC =
awrt
or
BAC =
awrt 0.88
(0.8785…)
9
1
π
3 3 8 sin(
"0.88")
( 20.4896....)
Area ABC
=
×
××
π
−
−
=
2
6
Total area = "18.8" "3.90" "20.5"
+
+
=
awrt 43
(m2)
If lengths or areas are found in other parts then credit can be awarded for these as long as
they are referred to or used in the relevant part.
(a)(i)
B1
Finds the correct angle for the sector AOBX. They may find the minor sector first and then
subtract from the area of the whole circle so may be implied from later work. They may also
4
π
work in degrees. Sight of
θ=
on the diagram or within their working scores this mark.
3
1
2
π
4
π
M1
States or uses
2 r θ with
2
r =
3
and
θ=
or
θ=
or the equivalent method working in
3
3
degrees. May be implied by the correct answer, expression or awrt 9.42 or awrt 18.8
A1
6π (m2) cao must be exact. Isw after a correct answer.
(ii)
2
π
4
π
M1
States or uses rθ with
r =
3
and
θ=
or
θ=
or the equivalent method working in
3
3
degrees. The addition of two radii to find the perimeter is not required for this mark.
May be implied by the correct answer, expression or awrt 6.28 or awrt 12.6
A1
4
π+ (m) cao must be exact. Isw after a correct answer.
6
(b)
States or uses 1
2
M1
sin
C with
=
=
3,
θ
=
π
(or may work in degrees). Alternatively,
2 ab
a
b
3
they may split the isosceles triangle into two right angled triangles including finding AB.
Score for the overall method or awrt 3.90
9 3
A1
(m2) or exact equivalent. Isw after a correct answer
(c)
M1
Correct method (which may be seen in earlier work but referred to in (c)) by for example
2
using the cosine rule with
a
=
b
=
3,
θ
=
"
π
"
3
π
splitting the isosceles triangle into two right angled triangles eg 2 3 sin
××
3
AB
3
using the sine rule such that
...
=
AB
=
2
π
π
sin
sin
3
6
A1*
3 3 (m) with no errors in their calculations. Minimum expected to see is a simplified
AB ) which is not 3 3 . Eg via the cosine rule this would be either
expression for AB (or
2
3sin(
2
)
π
AB =
2
27
or
AB =
27
via the sine rule eg
3
or via splitting the triangle up into
sin( )
π
6
π
two right angle triangles 2 3 sin
××
oe. Accept alternative labelling for AB eg x
3
You do not need to see the exact values for the the trig functions within their working.
(d)
Beware there are many methods with incorrect working leading to 43 so you will need
to check carefully that their method is sound.
M1
Attempts to use the sine rule to find sin BAC or angle BAC. Award for the appropriate
lengths and angles in the correct positions within a correct equation. Do not be concerned
with the mechanics of the rearrangement, although it must be a solvable equation.
Alternatively attempts to solve a quadratic in AC using the cosine rule.
π
(3 3)
2
=
AC
2
+
8
2
−×
2
AC
××
8 cos
AC
=
... (
=
11
+
4 3
=
awrt 10.2)
6
A1
Correct value for sin BAC , angle BAC (awrt 0.88 or 50.3 in degrees) or AC
M1
Correct method to find the area of triangle ABC using either their angle ABC from
π
π−
−
"
"
or their length AC. May work in degrees.
BAC
6
dM1
Adds their (a)(i) + their (b) + their triangle ABC together to the find area of the pond. It is
dependent on all of the previous method marks in (d) only
41
oe
A1
awrt 43 (m2) from a correct method. Also allow the exact answer
6
π+
2 11
+
3
4
Alt (d) Forms two right angled triangles with the perpendicular to side AC from point B
m2
m2
Y
π
M1
Height of the triangle ABC is BY = 8sin 6
A1
BY =
4
may be implied
M1
Attempts to find the area of triangle ABC. Usually this is by attempting to find length AC
(
)
2
(
)
π
=
split into AY and YC eg
AY =
3 3
−
"4"
2
=
11
and length YC =8cos
4 3
6
1
(
)
Area
=
×
11
+
4 3
×. Score for the overall method condoning slips.
4
ABC
2
Alternatively, they may attempt the cosine rule to find AC
Area
ABC
=
...
8(b)[2]
1
2
9 3
×
3
2
×
sin
π
=
(m2)
2
3
4
8(c)[2]
2
Eg
2
=
3
2
+
3
2
−×××
2 3 3 cos
π
2
=
27
=
3 3
(m) *
AB
AB
AB
3
8(d)[5]
π
sin
sin
BAC
6
=
sin
BAC
=
.... or
BAC
=
...
8
3 3
4 3
sin
BAC =
awrt
or
BAC =
awrt 0.88
(0.8785…)
9
1
π
3 3 8 sin(
"0.88")
( 20.4896....)
Area ABC
=
×
××
π
−
−
=
2
6
Total area = "18.8" "3.90" "20.5"
+
+
=
awrt 43
(m2)
If lengths or areas are found in other parts then credit can be awarded for these as long as
they are referred to or used in the relevant part.
(a)(i)
B1
Finds the correct angle for the sector AOBX. They may find the minor sector first and then
subtract from the area of the whole circle so may be implied from later work. They may also
4
π
work in degrees. Sight of
θ=
on the diagram or within their working scores this mark.
3
1
2
π
4
π
M1
States or uses
2 r θ with
2
r =
3
and
θ=
or
θ=
or the equivalent method working in
3
3
degrees. May be implied by the correct answer, expression or awrt 9.42 or awrt 18.8
A1
6π (m2) cao must be exact. Isw after a correct answer.
(ii)
2
π
4
π
M1
States or uses rθ with
r =
3
and
θ=
or
θ=
or the equivalent method working in
3
3
degrees. The addition of two radii to find the perimeter is not required for this mark.
May be implied by the correct answer, expression or awrt 6.28 or awrt 12.6
A1
4
π+ (m) cao must be exact. Isw after a correct answer.
6
(b)
States or uses 1
2
M1
sin
C with
=
=
3,
θ
=
π
(or may work in degrees). Alternatively,
2 ab
a
b
3
they may split the isosceles triangle into two right angled triangles including finding AB.
Score for the overall method or awrt 3.90
9 3
A1
(m2) or exact equivalent. Isw after a correct answer
(c)
M1
Correct method (which may be seen in earlier work but referred to in (c)) by for example
2
using the cosine rule with
a
=
b
=
3,
θ
=
"
π
"
3
π
splitting the isosceles triangle into two right angled triangles eg 2 3 sin
××
3
AB
3
using the sine rule such that
...
=
AB
=
2
π
π
sin
sin
3
6
A1*
3 3 (m) with no errors in their calculations. Minimum expected to see is a simplified
AB ) which is not 3 3 . Eg via the cosine rule this would be either
expression for AB (or
2
3sin(
2
)
π
AB =
2
27
or
AB =
27
via the sine rule eg
3
or via splitting the triangle up into
sin( )
π
6
π
two right angle triangles 2 3 sin
××
oe. Accept alternative labelling for AB eg x
3
You do not need to see the exact values for the the trig functions within their working.
(d)
Beware there are many methods with incorrect working leading to 43 so you will need
to check carefully that their method is sound.
M1
Attempts to use the sine rule to find sin BAC or angle BAC. Award for the appropriate
lengths and angles in the correct positions within a correct equation. Do not be concerned
with the mechanics of the rearrangement, although it must be a solvable equation.
Alternatively attempts to solve a quadratic in AC using the cosine rule.
π
(3 3)
2
=
AC
2
+
8
2
−×
2
AC
××
8 cos
AC
=
... (
=
11
+
4 3
=
awrt 10.2)
6
A1
Correct value for sin BAC , angle BAC (awrt 0.88 or 50.3 in degrees) or AC
M1
Correct method to find the area of triangle ABC using either their angle ABC from
π
π−
−
"
"
or their length AC. May work in degrees.
BAC
6
dM1
Adds their (a)(i) + their (b) + their triangle ABC together to the find area of the pond. It is
dependent on all of the previous method marks in (d) only
41
oe
A1
awrt 43 (m2) from a correct method. Also allow the exact answer
6
π+
2 11
+
3
4
Alt (d) Forms two right angled triangles with the perpendicular to side AC from point B
m2
m2
Y
π
M1
Height of the triangle ABC is BY = 8sin 6
A1
BY =
4
may be implied
M1
Attempts to find the area of triangle ABC. Usually this is by attempting to find length AC
(
)
2
(
)
π
=
split into AY and YC eg
AY =
3 3
−
"4"
2
=
11
and length YC =8cos
4 3
6
1
(
)
Area
=
×
11
+
4 3
×. Score for the overall method condoning slips.
4
ABC
2
Alternatively, they may attempt the cosine rule to find AC
Area
ABC
=
...

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