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ACTIVITIES
GUIDED ACTIVITY
628
A rectangle
ABCD
whose perimeter,
denoted by 2
p
, is
28
5
r
is inscribed in a circle
of radius
r
. Find the length
AB
and the
width
BC
of the rectangle, both as functions
of
r
.
O
A
D
B
C
OC
¼
r
2
p ABCD
ð
Þ ¼
28
5
r
AB
¼
?
BC
¼
?
The figure shows a rectangle
ABCD
inscribed in a circle of radius
r
. The diagonal
AC
divides the
rectangle into two congruent right triangles, both inscribed in a semicircle of radius
r
.
Focus on the triangle
ACD
. Considering what we know from the text, we can write:
AC
¼
2
r
;
AD
þ
DC
¼
1
2
28
5
r
¼
14
5
r
If we assume:
DC
¼
x
we automatically have:
AD
¼
14
5
r x
:
Now, using Pythagoras’ theorem on the triangle
ACD
, we obtain the resolvent equation. We
have:
AD
2
þ
DC
2
¼
AC
2
and so:
14
5
r x
2
þ
x
2
¼ ð
2
r
Þ
2
:
Now, the last equation is equivalent to the following: 25
x
2
70
rx
þ
48
r
2
¼
0
:
It has two real solutions:
x
1
;
2
¼
35
r
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1225
r
2
1200
r
2
p
25
¼
35
r
5
r
25
that is:
x
1
¼
6
5
r
and
x
2
¼
8
5
r
.
The presence of two different solutions means that we have two different rectangles satisfying
the given conditions, as we can observe in the following figures:
O
A
D
B
C
r
DC
¼
6
5
r
O
A
D
B
C
r
DC
¼
8
5
r
Do the following activity
629
A rectangle
ABCD
whose perimeter, denoted by 2
p
, is 70
a
is inscribed in a circle of radius
25
2
a
. Find the length
AB
and the width
BC
of the rectangle and its area.
½
AB
¼
20
a
,
CD
¼
15
a
or
AB
¼
15
a
,
CD
¼
20
a
;
A
¼
300
a
Unita` 1
Equazioni di grado superiore al 1
o
79