The perimeter of a rectangle is 2+2=
2
L
+
2
W
=
P
. The diagonals can be represented as 2+2=2
L
2
+
W
2
=
D
2
. Plugging in, we have:
2+2=42⟹+=21
2
L
+
2
W
=
42
⟹
L
+
W
=
21
2+2=152=225
L
2
+
W
2
=
15
2
=
225
We can square the first equation and subtract the second:
+=21
L
+
W
=
21
(+)2=212
(
L
+
W
)
2
=
21
2
2+2+2=441
L
2
+
2
L
W
+
W
2
=
441
2+2+2−(2+2)=441−225
L
2
+
2
L
W
+
W
2
−
(
L
2
+
W
2
)
=
441
−
225
2=216
2
L
W
=
216
It doesn’t tell us much, but we can now more easily substitute and solve for
L
.
=21−
W
=
21
−
L
& =108
L
W
=
108
(21−)=108
L
(
21
−
L
)
=
108
21−2=108
21
L
−
L
2
=
108
2−21+108=0
L
2
−
21
L
+
108
=
0
We can use the quadratic formula
=−±2−4√2
x
=
−
b
±
b
2
−
4
a
c
2
a
=21±441−432√2=21±9√2=21±32
L
=
21
±
441
−
432
2
=
21
±
9
2
=
21
±
3
2
=21+32=242=12
L
=
21
+
3
2
=
24
2
=
12
=21−32=182=9
L
=
21
−
3
2
=
18
2
=
9
We now know =9
L
=
9
or =12
L
=
12
. From this, we can calculate
W
by plugging
L
into =21−
W
=
21
−
L
.
=21−9=12
W
=
21
−
9
=
12
=21−12=9
W
=
21
−
12
=
9
We have two solutions:
1)=9
1
)
L
=
9
& =12
W
=
12
2)=12 2)L=12 &
=9 W=9
Both of which are the same, though. Our rectangle has side lengths of 9
9
, 12
12
, 9
9
, & 12
12
.