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### Step-by-Step Solution
1. Understanding the Shape:
The shape described is a rhombus, which consists of four congruent triangles. A rhombus has four equal sides, and its diagonals bisect each other at right angles (90 degrees).
2. Given Information:
- The side length of the rhombus is \(1\) unit.
- One of the diagonals is equal to the side length of the rhombus, so the length of this diagonal is also \(1\) unit.
3. Definition of Diagonals in a Rhombus:
A rhombus has two diagonals that intersect at right angles and bisect each other. This means each diagonal is divided into two equal segments by the point where they intersect.
4. Notation:
- Let \(d_1\) be the length of the known diagonal.
- Let \(d_2\) be the length of the unknown diagonal.
5. Known Diagonal:
Since one of the diagonals, \(d_1\), is equal to the side length of the rhombus, we have:
[tex]\[ d_1 = 1 \text{ unit} \][/tex]
6. Using the Pythagorean Theorem:
When diagonals of a rhombus intersect, they form four right-angled triangles with half of the diagonals as their legs. Each right-angled triangle will have legs of length \( \frac{d_1}{2} \) and \( \frac{d_2}{2} \), and the hypotenuse is the side length of the rhombus.
According to the Pythagorean theorem, for each right-angled triangle:
[tex]\[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = \text{side length}^2 \][/tex]
7. Substituting Known Values:
Substituting \(d_1 = 1\) and side length = 1:
[tex]\[ \left(\frac{1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 1^2 \][/tex]
8. Solving for \(d_2\):
Simplify the equation:
[tex]\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
[tex]\[ \frac{1}{4} + \left(\frac{d_2}{2}\right)^2 = 1 \][/tex]
Subtract \(\frac{1}{4}\) from both sides:
[tex]\[ \left(\frac{d_2}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \][/tex]
Now, solve for \(\left(\frac{d_2}{2}\right)\):
[tex]\[ \frac{d_2}{2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
9. Finding \(d_2\):
Multiply both sides by 2 to get \(d_2\):
[tex]\[ d_2 = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.224744871 \][/tex]
### Final Answer:
The side length of the rhombus is \(1\) unit, the known diagonal \(d_1\) is \(1\) unit, and the length of the unknown diagonal \(d_2\) is approximately \(1.224744871\) units.
Thus, the correct values are:
[tex]\[ \text{Side Length} = 1 \text{ unit} \][/tex]
[tex]\[ d_1 = 1 \text{ unit} \][/tex]
[tex]\[ d_2 \approx 1.224744871 \text{ units} \][/tex]
### Step-by-Step Solution
1. Understanding the Shape:
The shape described is a rhombus, which consists of four congruent triangles. A rhombus has four equal sides, and its diagonals bisect each other at right angles (90 degrees).
2. Given Information:
- The side length of the rhombus is \(1\) unit.
- One of the diagonals is equal to the side length of the rhombus, so the length of this diagonal is also \(1\) unit.
3. Definition of Diagonals in a Rhombus:
A rhombus has two diagonals that intersect at right angles and bisect each other. This means each diagonal is divided into two equal segments by the point where they intersect.
4. Notation:
- Let \(d_1\) be the length of the known diagonal.
- Let \(d_2\) be the length of the unknown diagonal.
5. Known Diagonal:
Since one of the diagonals, \(d_1\), is equal to the side length of the rhombus, we have:
[tex]\[ d_1 = 1 \text{ unit} \][/tex]
6. Using the Pythagorean Theorem:
When diagonals of a rhombus intersect, they form four right-angled triangles with half of the diagonals as their legs. Each right-angled triangle will have legs of length \( \frac{d_1}{2} \) and \( \frac{d_2}{2} \), and the hypotenuse is the side length of the rhombus.
According to the Pythagorean theorem, for each right-angled triangle:
[tex]\[ \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = \text{side length}^2 \][/tex]
7. Substituting Known Values:
Substituting \(d_1 = 1\) and side length = 1:
[tex]\[ \left(\frac{1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2 = 1^2 \][/tex]
8. Solving for \(d_2\):
Simplify the equation:
[tex]\[ \left(\frac{1}{2}\right)^2 = \frac{1}{4} \][/tex]
[tex]\[ \frac{1}{4} + \left(\frac{d_2}{2}\right)^2 = 1 \][/tex]
Subtract \(\frac{1}{4}\) from both sides:
[tex]\[ \left(\frac{d_2}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \][/tex]
Now, solve for \(\left(\frac{d_2}{2}\right)\):
[tex]\[ \frac{d_2}{2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]
9. Finding \(d_2\):
Multiply both sides by 2 to get \(d_2\):
[tex]\[ d_2 = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.224744871 \][/tex]
### Final Answer:
The side length of the rhombus is \(1\) unit, the known diagonal \(d_1\) is \(1\) unit, and the length of the unknown diagonal \(d_2\) is approximately \(1.224744871\) units.
Thus, the correct values are:
[tex]\[ \text{Side Length} = 1 \text{ unit} \][/tex]
[tex]\[ d_1 = 1 \text{ unit} \][/tex]
[tex]\[ d_2 \approx 1.224744871 \text{ units} \][/tex]
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