MEI Structured Mathematics
Practice Comprehension Question - 4
(Concepts for Advanced Mathematics, C2)
The 50p coin.
Feel a 50p coin. It has 7 edges but they are not straight lines; they are arcs of circles.
The coin is made this way so that it has constant width. This is particularly important when
such a coin is used in a coin machine (such as a ticket machine at train stations or car parks).
Obviously a circular coin will have constant width, but a coin made in the way described
below will also have constant width.
What is the percentage increase in the amount of metal needed to make the coin like that,
rather than a coin which is a regular heptagon?
5
To answer this question, start by looking at the regular heptagon in Fig. 1.
P
Q
r
θ
V
R
O
U
S
T
.
Fig.1
360D
.
7
Likewise, the lengths OP, OQ, OR …. are all the same and equal to the radius of the circle
that circumscribes the regular heptagon. In this article this radius is denoted by r.
The angles POQ, QOR,.. are all equal and are therefore all equal to θ =
10
To find the area of the heptagon, we first find the area of the triangle POQ.
There are several formulae for finding the area of a triangle when one angle is not a right
angle. One of them is
Area =
15
1
absinC
2
Applying this formula to the triangle POQ gives
1 2
r sin θ , and it follows that the area of the
2
heptagon is approximately 2.73641r2.
© MEI 2005
Practice Comprehension Question - 4
Page 1
Now for the 50p coin.
P
Q
Each of the edges is not a straight line, but the arc of a
circle.
20
The centre of this circle is the vertex opposite the edge,
so T is the centre of the arc PQ.
Fig. 2
The region enclosed in the diagram is a sector of a circle
with centre T and radius TP.
T
25
In order to find the area of this sector we need the angle PTQ and the radius TP ( = TQ).
Because O is the centre of the circumscribing circle of the heptagon,
Q
P
1
1 360D 180D
Angle PTQ = × Angle POQ = ×
=
.
2
2
7
7
Therefore Angle PTO =
1
1 180D 90D
× Angle PTQ = ×
=
.
2
2
7
7
r
r
O
r
Triangle POT is isosceles, with PO = OT = r.
Hence PT = 2rcos
90D
7
30
T
Fig. 3
angle PTQ
360D
2 180D
⎛
90D ⎞
7
= π ⎜ 2rcos
⎟ ×
D
7
360
⎝
⎠
D
2π 2 2 90
r cos
=
7
7
So the area of the sector is = π PT 2 ×
We are now in a position to find the extra metal in the 50p coin by considering the shaded
area in Fig. 4. This area is the difference in the area of the sector PTQ and the triangle PTQ.
© MEI 2005
Practice Comprehension Question - 4
35
Page 2
1
absinC
2
1
180D
= PT 2 sin
2
7
The area of triangle PTQ =
2
1⎛
90D ⎞
180D
= ⎜ 2r cos
sin
⎟
2⎝
7 ⎠
7
P
Q
90D
180D
= 2r cos
sin
7
7
2
2
40
Shaded area = Area of sector PTQ − Area of Triangle PTQ
⎛
90D ⎞ ⎛ 2π
180D ⎞
= ⎜ r 2 cos 2
−
2sin
⎟⎜
⎟
7 ⎠⎝ 7
7 ⎠
⎝
= 0.028353r 2
Fig. 4
T
So the area of the 50p coin is 2.7364r2 + 7× 0.028353r2 = 2.934871r2.
The percentage increase in the amount of metal needed is therefore approximately 7.25%.
© MEI 2005
Practice Comprehension Question - 4
Page 3
Questions:
1
Using the standard formula for the area of a triangle i.e.
of a triangle Area =
1
bh , prove the formula for the area
2
1
absinC , given in line 15.
2
[2]
2
Show that the area of the regular heptagon is 2.73641r2, as stated in line 17.
[3]
3
State the circle theorem being used in line 27.
[1]
90D
PT = 2rcos
7
4
Derive the formula in line 30:
5
Justify the last statement that the increase in metal is 7.25%
[3]
6
In 1997 the size of the 50p coin was reduced in size while remaining similar in shape.
Explain why the final answer on line 45 remains unaltered.
[1]
A new £5 coin is to be designed so that it has a constant width and is based on a regular
pentagon. Find the percentage increase in the amount of metal needed to make this coin
rather than a coin which is a regular pentagon?
[6]
7
© MEI 2005
Practice Comprehension Question - 4
[2]
Page 4
Answers.
1
A
h
B
1
ah where h = b sin C
2
b ⇒ Area = 1 ab sin C
2
Area =
Area OPQ =
A1
C
a
2
M1
2
1 2
r sinθ
2
M1
1
7
. ⇒ Area of heptagon = 7 × r 2sinθ = r 2 sin 51.43D
2
2
2
= 2.73641r
M1
A1
3
Angle at centre = 2 × angle at circumference.
B1
4
1
PT = r cos PTO
2
90D
⇒ PT = 2r cos
7
3
1
5
M1
A1
2
2
Area of 50p coin = 2.93487r .
Area of heptagon = 2.73641r2
2.93487r 2 − 2.73641r 2
⇒ Percentage increase =
× 100
2.73641r 2
0.19846r 2
=
×100
2.73641r 2
= 7.25
M1
A1
A1
3
6
Final answer is independent of r.
B1
1
© MEI 2005
Practice Comprehension Question - 4
Page 5
7
Angle POQ =
360D
= 72
5
B1
1 2
r sin 72 = 0.47553r 2
2
⇒ Area pentagon = 5 × 0.47553r 2 = 2.3776r 2
1
1
Angle PTQ = × 72 = 36 ⇒ Angle PTO = × 36 = 18D
2
2
D
⇒ PT = 2r cos18
⇒ Area Triangle POQ =
(
)
(
)
1
2r cos18D
2
⇒ Area Triangle PQT =
and Area Sector PTQ = π 2r cos18D
(
2
2
M1
A1
M1
sin 36D = 1.0633r 2
36D
= 1.13663r 2
D
360
A1
)
⇒ Area of extra bits = 5 1.13663r 2 − 1.0633r 2 = 0.3666r 2
Percentage increase =
© MEI 2005
0.3666r 2
× 100
2.3776r 2
= 15.4%
Practice Comprehension Question - 4
A1
6
Page 6