Certainly! Let's work through the problem step-by-step.
### Step-by-Step Solution:
1. Understanding the Problem:
- We have a rhombus formed by four congruent triangles.
- One of the diagonals of the rhombus is equal to the side length of the rhombus.
2. Labeling:
- Let the side length of the rhombus be denoted as [tex]\( a \)[/tex].
- The diagonals of the rhombus intersect at right angles and split the rhombus into four right triangles, each being congruent.
3. Properties of Rhombus and Diagonals:
- Diagonal [tex]\( D_1 \)[/tex] is given as equal to the side length of the rhombus, so [tex]\( D_1 = a \)[/tex].
- Let the other diagonal be denoted as [tex]\( D_2 \)[/tex].
- These diagonals intersect at right angles, breaking the rhombus into four right triangles.
4. Using the Pythagorean Theorem:
- In one of the right triangles, [tex]\( a \)[/tex] is the hypotenuse.
- The legs of this right triangle are half the lengths of the diagonals, i.e., [tex]\( D_1/2 \)[/tex] and [tex]\( D_2/2 \)[/tex].
- Therefore, the Pythagorean theorem can be applied to this right triangle:
[tex]\[
a^2 = \left( \frac{D_1}{2} \right)^2 + \left( \frac{D_2}{2} \right)^2
\][/tex]
- Substituting [tex]\( D_1 = a \)[/tex]:
[tex]\[
a^2 = \left( \frac{a}{2} \right)^2 + \left( \frac{D_2}{2} \right)^2
\][/tex]
- Simplifying, we get:
[tex]\[
a^2 = \frac{a^2}{4} + \frac{D_2^2}{4}
\][/tex]
[tex]\[
4a^2 = a^2 + D_2^2
\][/tex]
[tex]\[
3a^2 = D_2^2
\][/tex]
[tex]\[
D_2 = a \sqrt{3}
\][/tex]
5. Conclusion:
- The side length of the rhombus is [tex]\( a \)[/tex].
- The length of the shorter diagonal is [tex]\( D_1 = a \)[/tex].
- The length of the longer diagonal is [tex]\( D_2 = a \sqrt{3} \)[/tex].
Thus, if we consider [tex]\( a = 1 \)[/tex] unit for simplicity:
- The shorter diagonal is [tex]\( 1 \)[/tex] unit.
- The longer diagonal is approximately [tex]\( 1.732 \)[/tex] units.
In conclusion, the side length of the rhombus is 1 unit, the shorter diagonal is 1 unit, and the longer diagonal is approximately 1.732 units.
