Let's analyze and solve the problems in this second image.

Observations:

This task contains various exercises related to the perimeter of rectangles, involving equations to solve for the length ($a$) and width ($b$) based on given conditions.

Problem 1:

• Given: Perimeter $u = 170$, Width ($b$) = $44$
• Find: Length ($a$)

The perimeter formula is: $u = 2a + 2b$

Substitute the known values: $170 = 2a + 2(44)$ $170 = 2a + 88$ $2a = 170 - 88 = 82$ $a = \frac{82}{2} = 41$

Solution:

• Length ($a$) = $41$
• Width ($b$) = $44$

Problem 2a:

• Given: Perimeter $u = 70$, Length ($a$) = $x + 5$, Width ($b$) = $x$
• Find: $x$, Length, and Width

Using the perimeter formula: $u = 2a + 2b$

Substitute the given expressions: $70 = 2(x + 5) + 2x$ $70 = 2x + 10 + 2x$ $70 = 4x + 10$ $4x = 70 - 10 = 60$ $x = \frac{60}{4} = 15$

Solution:

• Width ($x$) = $15$
• Length ($x + 5$) = $20$

Problem 2b:

• Given: Perimeter $u = 50$, Length ($a$) = $x - 7$, Width ($b$) = $x$
• Find: $x$, Length, and Width

Using the perimeter formula: $u = 2a + 2b$

Substitute the given expressions: $50 = 2(x - 7) + 2x$ $50 = 2x - 14 + 2x$ $50 = 4x - 14$ $4x = 50 + 14 = 64$ $x = \frac{64}{4} = 16$

Solution:

• Width ($x$) = $16$
• Length ($x - 7$) = $9$

Problem 2c:

This is identical to Task c from the previous image:

• Given: Perimeter $u = 42$, Length ($a$) = $2x$, Width ($b$) = $x$
• Solution: As solved earlier:
  • Length = $14$
  • Width = $7$

Problem 2d:

This is identical to Task d from the previous image:

• Given: Perimeter $u = 24$, Length ($a$) = $x$, Width ($b$) = $\frac{1}{2}x$
• Solution: As solved earlier:
  • Length = $8$
  • Width = $4$

Final Summary of Solutions:

1. Problem 1: Length = $41$, Width = $44$
2. Problem 2a: Length = $20$, Width = $15$
3. Problem 2b: Length = $9$, Width = $16$
4. Problem 2c: Length = $14$, Width = $7$
5. Problem 2d: Length = $8$, Width = $4$

Would you like me to elaborate on any specific step or equation?

5 Related Questions:

1. How does changing the perimeter affect the length and width of a rectangle?
2. What happens if the width is double the length instead of half?
3. Can you derive the area of each rectangle based on these solutions?
4. How would you solve these problems if decimals were involved instead of integers?
5. What are some real-life applications of perimeter equations?

Tip:

To verify your solutions, always plug the values of length and width back into the perimeter formula. This ensures accuracy!