The Enclosed one
The figure above shows two circles , each of radius 1 cm passes through each center. A square is inscribed inside the region enclosed between these two circles. Find the side length of this square (in cm ).
Give your answer to 3 decimal places.
The answer is 0.822.
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2 solutions
Nice one! But the Diagram needs a little correction...
What is the error in diagram?
circles centers are not in the correct position.. ha ha! ;( just a joke.. but nice solution using CG .cheers!
yes, you are right. I'll correct it.
Now its perfect... Upvoted!
Square with EB as a diagonal is quarter of square ABCD.
Also F is mid point of BC, E of GF and OO'. And BF=EF= say x. OE=EO'=
2
1
.
Using Pythagoras Theorem in right triangle OBF.
O
B
2
=
O
F
2
+
B
F
2
,
⟹
1
2
=
(
2
1
+
x
)
2
+
x
∗
2
.
S
o
l
v
i
n
g
q
u
a
d
r
a
t
i
c
i
n
x
,
x
=
0
.
4
1
1
4
.
B
u
t
B
C
=
2
x
=
0
.
8
2
2
8
.
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Your approach is same as of mine.. only the difference is that you have used the upper portion of the square.. good solution though.. cheers!
nyc approach !
If you want the exact answer without using graphs:
Take the side of the square to be 'a'
BE² = EF² + FB²
BE = √(2*(a/(2))²) = a/(√2)
Triangle BEO has sides BE = a/(√2), B0 = 1 and EO = 0.5
Angle BEO = 90° + 45° = 135°
Using cosine rule on triangle BEO,
1² = 0.5² + (a/(√2))² - 2 * 0.5 * a/(√2) * cos(135)
1 = 1/4 + a²/2 - a/(√2) * - √(2)/ 2
1 = 1/4 + a²/2 + a/2
After some rearranging...
2a² + 2a - 3 = 0
a = (√(7)-1)/2 (= 0.8228...)