To find the smallest diameter of the cylindrical container that will fit over candy bar ABC, we need to analyze the dimensions given in the problem.
We know that segment AD is 4 cm. Since points E and F are midpoints (due to segments AE and EB being congruent, and BF and FC being congruent), we can denote segment AB and segment BC using the parameters derived from the triangle.
From the right angles given (∠AED, ∠ABF, and ∠DFC), we can deduce that triangle ABC forms a rectangle when combined with points D and F, giving us right angles at these points.
- Determine Lengths AB and AC Using AD:
Segment AD = 4 cm, which means that point D is 4 cm vertically. Since E is the midpoint of AB, and we have two equal segments AE and EB, each will then measure \( x \) cm.
If AE = EB = x, then we get:
- \( AB = AE + EB = x + x = 2x \).
- Determine Lengths BC:
Since point F is also the midpoint of BC, similar reasoning applies, leading to:
- \( BC = BF + FC = y + y = 2y \).
- Triangle Dimensions:
Since D is directly below C, the horizontal distance covered (base of triangle ABC) can be represented as:
- Base = AB + BC = \( 2x + 2y \).
- Resolving Variables:
Given that segment AE (4 cm) forms part of a right triangle (ABD), the lengths can be expressed as \(\sqrt{(AD^2 + AB^2)}\).
The height (AD) = 4 cm is vertical, and the length of AB (which we've expressed as \(2x\)) can be represented by the width of the candy bar.
- Finding Diameter:
To fit the entire candy bar within a cylindrical container, the width and height determine the bounds of the cylinder. Thus, height = vertical (4 cm) + \(y\) and width is \(2x + 2y\).
The smallest container must accommodate:
- Height: \(4 + y\)
- Width: \(2x + 2y\)
To account for a cylindrical container, we'd need to ensure that the cylinder's diameter is enough to cover the widest part of the candy, hence we need the maximum length dimension, given by \(D\).
Now, adding our right angle triangle dimensioning:
- Diameter = Maximum width = \(AD + AB + BC = 4 + 2x + 2y\).
On further calculations:
- If AD is the overall base then diameter = \(Width x Height,\)
- Assuming the candy does fit appropriately leading to a diameter resolution yielding minimal diameter wrapped around:
Finally, with proper studies of limits available (7 cm, 8 cm, 9 cm, 10 cm), and base rectangle allowance mixed with sides, we deduce the diameter must specifically accommodate the least presentation ensuring no outside projection fits:
Therefore the smallest diameter that will fit the candy bar would be: Answer: \(8 \text{ cm}\).
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