Let $ABC$ be a triangle of area $1$ with medians $s_a, s_b,s_c$. Show that there is a triangle whose sides are the same length as $s_a, s_b$, and $s_c$, and determine its area.
1. **Given Information and Setup:**
   - We are given a triangle \( \triangle ABC \) with area 1.
   - The medians of the triangle are denoted as \( s_a, s_b, s_c \).
   - We need to show that there exists a triangle whose sides are \( s_a, s_b, s_c \) and determine its area.

2. **Coordinates and Midpoints:**
   - Let the coordinates of the vertices be \( A(0,0) \), \( B(2,0) \), and \( C(\lambda,1) \).
   - The area of \( \triangle ABC \) is given by:
     \[
     \text{Area} = \frac{1}{2} \left| 0(0-1) + 2(1-0) + \lambda(0-0) \right| = 1
     \]
     Simplifying, we get:
     \[
     1 = \frac{1}{2} \left| 2 \right| \implies 2 = 2 \implies \lambda = 1
     \]
     So, the coordinates of \( C \) are \( (1,1) \).

3. **Finding the Midpoints:**
   - Midpoint \( E \) of \( BC \):
     \[
     E \left( \frac{2+1}{2}, \frac{0+1}{2} \right) = E \left( \frac{3}{2}, \frac{1}{2} \right)
     \]
   - Midpoint \( D \) of \( AC \):
     \[
     D \left( \frac{0+1}{2}, \frac{0+1}{2} \right) = D \left( \frac{1}{2}, \frac{1}{2} \right)
     \]
   - Midpoint \( F \) of \( AB \):
     \[
     F \left( \frac{0+2}{2}, \frac{0+0}{2} \right) = F(1,0)
     \]

4. **Calculating the Lengths of the Medians:**
   - Length of median \( AE \):
     \[
     AE = \sqrt{ \left( \frac{3}{2} \right)^2 + \left( \frac{1}{2} \right)^2 } = \sqrt{ \frac{9}{4} + \frac{1}{4} } = \sqrt{ \frac{10}{4} } = \frac{\sqrt{10}}{2}
     \]
   - Length of median \( BD \):
     \[
     BD = \sqrt{ \left( 2 - \frac{1}{2} \right)^2 + \left( 0 - \frac{1}{2} \right)^2 } = \sqrt{ \left( \frac{3}{2} \right)^2 + \left( -\frac{1}{2} \right)^2 } = \sqrt{ \frac{9}{4} + \frac{1}{4} } = \sqrt{ \frac{10}{4} } = \frac{\sqrt{10}}{2}
     \]
   - Length of median \( CF \):
     \[
     CF = \sqrt{ (1-1)^2 + (1-0)^2 } = \sqrt{ 0 + 1 } = 1
     \]

5. **Forming the New Triangle:**
   - The sides of the new triangle are \( s_a = \frac{\sqrt{10}}{2} \), \( s_b = \frac{\sqrt{10}}{2} \), and \( s_c = 1 \).

6. **Area of the New Triangle:**
   - The area of a triangle with medians \( s_a, s_b, s_c \) can be found using the formula:
     \[
     \text{Area} = \frac{4}{3} \times \text{Area of the original triangle}
     \]
     Given that the area of \( \triangle ABC \) is 1, the area of the new triangle is:
     \[
     \text{Area} = \frac{4}{3} \times 1 = \frac{4}{3}
     \]

The final answer is \( \boxed{ \frac{4}{3} } \)