Let $ABC$ be a triangle with area $92$ square centimeters. Calculate the area of another triangle whose sides have the same lengths as the medians of triangle $ABC$.
1. **Understanding the Problem:**
   We are given a triangle \(ABC\) with an area of \(92\) square centimeters. We need to find the area of another triangle whose sides are the medians of triangle \(ABC\).

2. **Key Concept:**
   The area of a triangle formed by the medians of another triangle is \(\frac{3}{4}\) of the area of the original triangle. This is a well-known result in geometry.

3. **Proof of the Key Concept:**
   - Let \(D, E, F\) be the midpoints of sides \(BC, CA, AB\) respectively.
   - The medians of triangle \(ABC\) intersect at the centroid \(G\), which divides each median in the ratio \(2:1\).
   - Construct a point \(G'\) such that \(BF\) is parallel and equal to \(DG'\). Then \(DFG'C\) forms a parallelogram.
   - The diagonals of parallelogram \(DFG'C\) bisect each other, so \(H\) is the midpoint of both \(CF\) and \(DG'\).
   - Since \(BEGC\) is a parallelogram, \(\angle AEG = \angle ABC\), \(BC = EG\), and \(AE = EB\). Thus, triangles \(AEG\) and \(EBC\) are congruent, implying \(AG = EC\).
   - Therefore, triangle \(ADG\) has side lengths equal to the medians of triangle \(ABC\).
   - The area ratio \(\frac{[ADG]}{[ABC]} = \frac{2[ADH]}{2[ADC]} = \frac{AH}{AC} = \frac{3}{4}\).

4. **Calculation:**
   Given the area of triangle \(ABC\) is \(92\) square centimeters, the area of the triangle formed by the medians is:
   \[
   \text{Area of the new triangle} = \frac{3}{4} \times 92 = 69 \text{ square centimeters}
   \]

The final answer is \(\boxed{69}\).