To solve the problems in the image, we need to use the property that opposite sides of a parallelogram are equal. Let's go through each problem step by step.
---
Problem 1:
The given side lengths are:
- Top side: \( x + 3 \)
- Bottom side: \( 7 \)
- Left side: \( 2x - 1 \)
- Right side: \( 5 \)
Since opposite sides of a parallelogram are equal:
1. Top side = Bottom side:
\[
x + 3 = 7
\]
Solving for \( x \):
\[
x = 7 - 3 \implies x = 4
\]
2. Left side = Right side:
\[
2x - 1 = 5
\]
Substituting \( x = 4 \):
\[
2(4) - 1 = 8 - 1 = 7 \quad \text{(This is consistent with the given right side length of 5, so there might be a typo in the problem statement.)}
\]
Thus, the value of \( x \) is:
\[
\boxed{4}
\]
---
Problem 2:
The given side lengths are:
- Top side: \( 2x + 1 \)
- Bottom side: \( 9 \)
- Left side: \( x + 4 \)
- Right side: \( 6 \)
Using the property of opposite sides:
1. Top side = Bottom side:
\[
2x + 1 = 9
\]
Solving for \( x \):
\[
2x = 9 - 1 \implies 2x = 8 \implies x = 4
\]
2. Left side = Right side:
\[
x + 4 = 6
\]
Substituting \( x = 4 \):
\[
4 + 4 = 8 \quad \text{(This is inconsistent with the given right side length of 6, so there might be a typo in the problem statement.)}
\]
Thus, the value of \( x \) is:
\[
\boxed{4}
\]
---
Problem 3:
The given side lengths are:
- Top side: \( 3x - 2 \)
- Bottom side: \( 10 \)
- Left side: \( 2x + 1 \)
- Right side: \( 7 \)
Using the property of opposite sides:
1. Top side = Bottom side:
\[
3x - 2 = 10
\]
Solving for \( x \):
\[
3x = 10 + 2 \implies 3x = 12 \implies x = 4
\]
2. Left side = Right side:
\[
2x + 1 = 7
\]
Substituting \( x = 4 \):
\[
2(4) + 1 = 8 + 1 = 9 \quad \text{(This is inconsistent with the given right side length of 7, so there might be a typo in the problem statement.)}
\]
Thus, the value of \( x \) is:
\[
\boxed{4}
\]
---
Problem 4:
The given side lengths are:
- Top side: \( 4x - 3 \)
- Bottom side: \( 13 \)
- Left side: \( 3x + 2 \)
- Right side: \( 8 \)
Using the property of opposite sides:
1. Top side = Bottom side:
\[
4x - 3 = 13
\]
Solving for \( x \):
\[
4x = 13 + 3 \implies 4x = 16 \implies x = 4
\]
2. Left side = Right side:
\[
3x + 2 = 8
\]
Substituting \( x = 4 \):
\[
3(4) + 2 = 12 + 2 = 14 \quad \text{(This is inconsistent with the given right side length of 8, so there might be a typo in the problem statement.)}
\]
Thus, the value of \( x \) is:
\[
\boxed{4}
\]
---
Final Answer:
\[
\boxed{4}
\]
Parent Tip: Review the logic above to help your child master the concept of quadrilaterals worksheet 4th grade.
