Rectangles Squared
Two congruent rectangles of dimensions 1 × x are placed inside a square of side length 1. One of them is vertical, and the other is inclined.
Find the side length x to 3 decimal places.
The answer is 0.267.
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2 solutions
Nice problem and solution. One typo to point out; I think that you meant that cos ( θ ) = 2 1 , giving us θ = 3 π radians.
Nice simple solution. up voted. I post my solution just as another solution. Not a better one.
It would be interesting to know how you solved the equations. I would do it as under.
x
C
o
s
θ
+
S
i
n
θ
=
1
.
.
.
.
.
.
.
.
.
.
.
.
.
(
A
)
x
+
x
S
i
n
θ
+
C
o
s
θ
=
1
.
.
.
.
.
.
.
.
.
(
B
)
F
r
o
m
(
A
)
x
=
C
o
s
θ
1
−
S
i
n
θ
F
r
o
m
(
B
)
x
=
1
+
S
i
n
θ
1
−
C
o
s
θ
∴
(
1
−
S
i
n
θ
)
(
1
+
S
i
n
θ
)
=
C
o
s
θ
(
1
−
C
o
s
θ
)
∴
C
o
s
2
θ
=
C
o
s
θ
(
1
−
C
o
s
θ
)
,
C
o
s
θ
=
0
.
⟹
C
o
s
θ
=
1
−
C
o
s
θ
.
⟹
C
o
s
θ
=
2
1
.
Let θ be the angle of inclination relative to the horizontal of the second rectangle.
From that we have that the sides of the square become
x
cos
θ
+
sin
θ
and
x
+
x
sin
θ
+
cos
θ
which constitute a system of equations because we have the side of the square,
x
cos
θ
+
sin
θ
=
1
x
+
x
sin
θ
+
cos
θ
=
1
Solving it would lead to
cos
θ
=
2
1
and
x
=
2
−
3
, using three decimal places the answer is
2
−
1
,
7
3
2
0
5
0
8
0
7
=
0
.
2
6
7