What is the area of the inscribed quadrilateral
ABCD?
A
28
B
36
C
42
D
48
E
56
Solution:
A
The graph cuts the
y
-axis at points where
0
=
x
, and hence where
12
4
2
=
−
y
y
, that is, where
0
12
4
2
=
−
−
y
y
, that is,
0
)
6
)(
2
(
=
−
+
y
y
, giving
2
−
=
y
and
6
=
y
. Hence
A
is the point (0,6),
C
is the point
)
2
,
0
(
−
and so
AC
has length 8.
Similarly,
B
and
D
are points where
0
=
y
and hence
0
12
2
=
−
+
x
x
, that is,
0
)
3
)(
4
(
=
−
+
x
x
, giving
4
−
=
x
and
3
=
x
. So
B
is the point
)
0
,
3
(
,
D
is
the point
)
0
,
4
(
−
, and so
BD
has length 7. The area of the inscribed quadrilateral
ABCD
is therefore
BD
AC
.
2
1
28
)
7
8
(
2
1
=
×
=
.
Extension Problems
16.1
The solution above takes it for granted that the area of the quadrilateral
ABCD
is half the
product of the lengths of its diagonals. Give an example to show that this is
not true
for all
quadrilaterals.
16.2 What is the special property of
ABCD
that makes it correct that in this case the area is half
the product of the length of the diagonals? Give a proof to show that your answer is correct.
17.
The diagram shows a pattern found on a floor tile in the cathedral
in Spoleto, Umbria. A circle of radius 1 surrounds four quarter
circles, also of radius 1, which enclose a square. The pattern has
four axes of symmetry. What is the side length of the square?
A
2
1
B
2
2
−
C
3
1
D
2
1
E
1
2
−
Solution:
B
Let
O
be the centre of the circle, let
P
be one of the points where
two of the quarter circles meet, let
Q
be the centre of one of
these quarter circles, and let
R
be the vertex of the square that
lies on
OQ
, as shown.
Then in the triangle
OPQ
there is a right angle at
P
and
1
=
=
QP
OP
,
and therefore, by Pythagoras’ Theorem,
2
=
OQ
. Since
1
=
QR
, it follows that
1
2
−
=
OR
. It
follows that the diagonal of the square has length
)
1
2
(
2
−
. Using, Pythagoras’ Theorem again, it
follows that the side length of the square is
(
)
)
1
2
(
2
2
1
−
2
2
−
=
.
Extension Problem
17.1 In the proof above we claim, by Pythagoras’ Theorem, that if the diagonal of a square has
length
x
, then its side length is
x
2
1
. Show that this is indeed a consequence of Pythagoras’
Theorem.