Mathematics · Oct / Nov 2020 · Paper 11 · Question 3
YMA01
As printed in the paper
The question as text
3. A Diagram not drawn to scale 3 cm C α 5 cm O 3 cm B Figure 1 Figure 1 shows the design for a badge. The design consists of two congruent triangles, AOC and BOC, joined to a sector AOB of a circle centre O. Angle AOB = α AO = OB = 3 cm OC = 5 cm Given that the area of sector AOB is 7.2 cm2
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
1
1
×
3
2
×
α
=
7.2
⇒
α
=
...
or
×
3
2
×
1.6
=
7.2
⇒
α
=
1.6
2
2
*
α =
1.6
(b)(i)
1
(
)(
)
(
≈
134
°)
Angle
COA
=
2
π
−
1.6
=
2.34...
2
1
(
)
( 5.38...)
=
Area
COA =
× ×
5 3sin "2.34"
2
1
(
)
TotalArea
=
2
×
× ×
5 3sin "2.34"
+
7.2
2
= 18 (cm2) Awrt 18 (cm2) (Ans = 17.96)
(ii)
(
)
Arc
AB = ×
3 1.6
=
4.8
(
)
(
)
AC =
2
5
2
+
3
2
−× ×
2 5 3cos "2.34"
Total perimeter
2
5
2
3
2
2 5 3cos "2.34"
(
)
3 1.6
=
×
+
−× ×
+ ×
= Awrt 19.6 (cm)
Alt (b)(i)
AB =
2 3sin 0.8
×
3cos0.8
ON =
1
1
(
)
TotalArea
=
5
+
ON
×
AB
+
7.2
−
×
3cos0.8 2 3sin 0.8
× ×
2
2
= 18 (cm2) Awrt 18 (cm2) (Ans = 17.96)
Notes
(a)
M1 Uses a correct sector area formula and 7.2 to find the value for α. They should show the values
embedded in the equation and proceed to find a value for α .
Alternatively, substitutes in
1.6
into the area of a sector formula and achieves 7.2.
α =
1
A1* Correct proof starting with
3
2
7.2
and at least one intermediate line of working and no
×
×
α
=
2
1
7.2
errors. Eg
×
3
2
×
α
=
7.2
⇒
α
=
=
1.6
scores M1A1
2
4.5
Alternatively, they must conclude that
or if there is a preamble then there should be
α =
1.6
some form of completion which could be a tick, QED etc.
If they use a different variable such as θ they must state/link somewhere that
α =
1.6
(b)(i) Mark both (i) and (ii) together. If no angle calculation is seen then use what they think is
their angle COA in bi and bii. Beware of values on the diagram that may imply a method.
1 2
M1
(
)
Correct method for angle COA. Sight of awrt 2.34 is sufficient to score this mark
π −
1.6
2
and may be on the diagram. (May also be implied by 134°)
M1 Uses a correct method for the area of triangle COA or COB. It is sufficient to see the values
1
(
)
embedded in the expression such as
2 × ×
5 3sin "2.34"
( awrt5.38 / 5.39)
=
. Angle may be in
degrees. If they state 1
C oe but embed values as 1
2 abC condone as a slip for M1.
2 ab
sin
dM1 Fully correct strategy for the area. It is dependent on the previous method mark so allow if their
angle COA is incorrect. Look for 2×area of triangle COA 7.2
+
. Embedded values are sufficient.
A1
awrt 18 (cm2) Must come from a correct method
…………………………………………………………………………………………………………….
Alt b(i)
M1 Find the length AB
2 3sin 0.8
(awrt 4.30)
=
×
M1 Finds the length ON where N is the midpoint of AB (awrt 2.09)
dM1 Fully correct strategy for the area. Look for 7.2
+
area of triangle
ABC
−
area of triangle
A1
awrt 18 (cm2) Must come from a correct method
…………………………………………………………………………………………………………….
(ii)
B1
Correct expression or value for the arc length (4.8)
M1 Uses a correct method for AC2, AC, CB2 or CB. Embedded values in the associated formula is
sufficient or sight of awrt 54.9 or awrt 7.41 would imply this mark. (Angle in degrees
For this mark condone candidates confusing AC2/CB2 and AC/CB.
dM1
Total perimeter
2
5
2
3
2
2 5 3cos "2.34"
(
)
3 1.6
. It is dependent on the B and the
=
×
+
−× ×
+ ×
M marks and they must have remembered to square root AC2 or CB2.
A1
awrt 19.6 (cm) (Ans = 19.619) Must come from a correct method
3(b)[8]
1
1
×
3
2
×
α
=
7.2
⇒
α
=
...
or
×
3
2
×
1.6
=
7.2
⇒
α
=
1.6
2
2
*
α =
1.6
(b)(i)
1
(
)(
)
(
≈
134
°)
Angle
COA
=
2
π
−
1.6
=
2.34...
2
1
(
)
( 5.38...)
=
Area
COA =
× ×
5 3sin "2.34"
2
1
(
)
TotalArea
=
2
×
× ×
5 3sin "2.34"
+
7.2
2
= 18 (cm2) Awrt 18 (cm2) (Ans = 17.96)
(ii)
(
)
Arc
AB = ×
3 1.6
=
4.8
(
)
(
)
AC =
2
5
2
+
3
2
−× ×
2 5 3cos "2.34"
Total perimeter
2
5
2
3
2
2 5 3cos "2.34"
(
)
3 1.6
=
×
+
−× ×
+ ×
= Awrt 19.6 (cm)
Alt (b)(i)
AB =
2 3sin 0.8
×
3cos0.8
ON =
1
1
(
)
TotalArea
=
5
+
ON
×
AB
+
7.2
−
×
3cos0.8 2 3sin 0.8
× ×
2
2
= 18 (cm2) Awrt 18 (cm2) (Ans = 17.96)
Notes
(a)
M1 Uses a correct sector area formula and 7.2 to find the value for α. They should show the values
embedded in the equation and proceed to find a value for α .
Alternatively, substitutes in
1.6
into the area of a sector formula and achieves 7.2.
α =
1
A1* Correct proof starting with
3
2
7.2
and at least one intermediate line of working and no
×
×
α
=
2
1
7.2
errors. Eg
×
3
2
×
α
=
7.2
⇒
α
=
=
1.6
scores M1A1
2
4.5
Alternatively, they must conclude that
or if there is a preamble then there should be
α =
1.6
some form of completion which could be a tick, QED etc.
If they use a different variable such as θ they must state/link somewhere that
α =
1.6
(b)(i) Mark both (i) and (ii) together. If no angle calculation is seen then use what they think is
their angle COA in bi and bii. Beware of values on the diagram that may imply a method.
1 2
M1
(
)
Correct method for angle COA. Sight of awrt 2.34 is sufficient to score this mark
π −
1.6
2
and may be on the diagram. (May also be implied by 134°)
M1 Uses a correct method for the area of triangle COA or COB. It is sufficient to see the values
1
(
)
embedded in the expression such as
2 × ×
5 3sin "2.34"
( awrt5.38 / 5.39)
=
. Angle may be in
degrees. If they state 1
C oe but embed values as 1
2 abC condone as a slip for M1.
2 ab
sin
dM1 Fully correct strategy for the area. It is dependent on the previous method mark so allow if their
angle COA is incorrect. Look for 2×area of triangle COA 7.2
+
. Embedded values are sufficient.
A1
awrt 18 (cm2) Must come from a correct method
…………………………………………………………………………………………………………….
Alt b(i)
M1 Find the length AB
2 3sin 0.8
(awrt 4.30)
=
×
M1 Finds the length ON where N is the midpoint of AB (awrt 2.09)
dM1 Fully correct strategy for the area. Look for 7.2
+
area of triangle
ABC
−
area of triangle
A1
awrt 18 (cm2) Must come from a correct method
…………………………………………………………………………………………………………….
(ii)
B1
Correct expression or value for the arc length (4.8)
M1 Uses a correct method for AC2, AC, CB2 or CB. Embedded values in the associated formula is
sufficient or sight of awrt 54.9 or awrt 7.41 would imply this mark. (Angle in degrees
For this mark condone candidates confusing AC2/CB2 and AC/CB.
dM1
Total perimeter
2
5
2
3
2
2 5 3cos "2.34"
(
)
3 1.6
. It is dependent on the B and the
=
×
+
−× ×
+ ×
M marks and they must have remembered to square root AC2 or CB2.
A1
awrt 19.6 (cm) (Ans = 19.619) Must come from a correct method
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