To solve this problem, we need to follow these steps:
1. Calculate the area of the given parallelogram PQRS.
2. Determine the side length of the new parallelogram with one angle measuring [tex]\(30^\circ\)[/tex] that will match the area of parallelogram PQRS.
### Step 1: Calculate the area of parallelogram PQRS
Given:
- Side [tex]\(PQ = 4.2 \text{ cm}\)[/tex]
- Side [tex]\(QR = 6 \text{ cm}\)[/tex]
- Angle [tex]\(\angle PQR = 60^\circ\)[/tex]
The formula for the area [tex]\(A\)[/tex] of a parallelogram is given by:
[tex]\[ A = a \cdot b \cdot \sin(\theta) \][/tex]
Where:
- [tex]\(a\)[/tex] is the length of one side (PQ)
- [tex]\(b\)[/tex] is the length of the adjacent side (QR)
- [tex]\(\theta\)[/tex] is the angle between them
Substitute the given values:
[tex]\[ A = 4.2 \, \text{cm} \cdot 6 \, \text{cm} \cdot \sin(60^\circ) \][/tex]
Given the answer from the calculations:
[tex]\[ A = 21.823840175367856 \, \text{cm}^2 \][/tex]
### Step 2: Construct a parallelogram with one angle [tex]\(30^\circ\)[/tex]
We need a new parallelogram with:
- One side length [tex]\(b_{\text{new}} = QR = 6 \, \text{cm}\)[/tex]
- An angle of [tex]\(30^\circ\)[/tex]
- The same area as parallelogram PQRS (21.823840175367856 cm²)
Let the unknown side length of the new parallelogram be [tex]\(a_{\text{new}}\)[/tex].
Using the area formula again:
[tex]\[ \text{Area} = a_{\text{new}} \cdot b_{\text{new}} \cdot \sin(30^\circ) \][/tex]
Substitute the known values and solve for [tex]\(a_{\text{new}}\)[/tex]:
[tex]\[ 21.823840175367856 \, \text{cm}^2 = a_{\text{new}} \cdot 6 \, \text{cm} \cdot \sin(30^\circ) \][/tex]
We know [tex]\(\sin(30^\circ) = 0.5\)[/tex]:
[tex]\[ 21.823840175367856 \, \text{cm}^2 = a_{\text{new}} \cdot 6 \, \text{cm} \cdot 0.5 \][/tex]
[tex]\[ 21.823840175367856 \, \text{cm}^2 = a_{\text{new}} \cdot 3 \, \text{cm} \][/tex]
Now, solve for [tex]\(a_{\text{new}}\)[/tex]:
[tex]\[ a_{\text{new}} = \frac{21.823840175367856 \, \text{cm}^2}{3 \, \text{cm}} \][/tex]
[tex]\[ a_{\text{new}} = 7.274613391789287 \, \text{cm} \][/tex]
### Conclusion
The area of parallelogram PQRS is approximately 21.82 cm². For the new parallelogram with an angle of 30°, to have the same area, one side length should be approximately 7.27 cm while the other side remains 6 cm.
