Let's first draw the rhombus and label the diagonals:
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A
o
/ \
3.5 12
/ \
o----x----o
\ /
3.5 12
\ /
o B
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Let the rhombus be ABCD, with AB = BC = CD = DA. Let O be the center of the circle, which is also a vertex of the rhombus. Then, OA and OB are radii of the circle, and they are also perpendicular bisectors of sides AB and BC, respectively. Therefore, triangle AOB is a right triangle, and we can use the Pythagorean Theorem to find the length of OB:
OA = OB = OC = OD (since O is the center of the circle)
AB = BC = 12 (since 12 is the length of diagonal AC)
AO^2 = AB^2/4 + OB^2 (since AO and OB are the legs of right triangle AOB)
Substituting AB = 12 and simplifying, we get:
OB^2 = AO^2 - AB^2/4
= (3.5/2)^2 - 12^2/4
= 49/16 - 144/4
= 49/16 - 36
= 1/16
Taking the square root of both sides, we get:
OB = \sqrt{1/16} = 1/4
Now, the circle is tangent to sides AB and BC, so its diameter must be perpendicular to these sides. Therefore, the diameter of the circle is equal to the length of diagonal BD, which is the hypotenuse of right triangle AOB:
BD^2 = AB^2 + OB^2
= 12^2 + (1/4)^2
= 144 + 1/16
= 577/16
Taking the square root of both sides, we get:
BD = \sqrt{577}/4
Finally, the area of the circle is given by:
A = pi*(BD/2)^2
= pi*(\sqrt{577}/8)^2
= pi*577/64
Therefore, the exact value of the circle area is (577/64)*pi.