Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
### Part (a): Simplification
Let's simplify the problem given in part (b) step by step.
### Part (b): Solving the Problem
#### Step 1: Understanding the Problem
We are given a wire of length 38 cm that is bent into the shape of a rectangle. The length of the rectangle is 7 cm more than the width. We need to find:
1. The width and length of the rectangle.
2. The area of the rectangle.
3. Given farone is 720 metres, find half of the length of the rope.
#### Step 2: Representing the Variables
Let's denote:
- The width of the rectangle by [tex]\( w \)[/tex] cm.
- The length of the rectangle by [tex]\( l \)[/tex] cm.
According to the problem, the length [tex]\( l \)[/tex] is 7 cm more than the width [tex]\( w \)[/tex]:
[tex]\[ l = w + 7 \][/tex]
#### Step 3: Using Perimeter to Formulate Equations
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (l + w) \][/tex]
Given that the wire length (which is the perimeter) is 38 cm, we can write:
[tex]\[ 2 \times (l + w) = 38 \][/tex]
Substitute [tex]\( l = w + 7 \)[/tex] into the perimeter equation:
[tex]\[ 2 \times ((w + 7) + w) = 38 \][/tex]
Simplify and solve for [tex]\( w \)[/tex]:
[tex]\[ 2 \times (2w + 7) = 38 \][/tex]
[tex]\[ 4w + 14 = 38 \][/tex]
[tex]\[ 4w = 24 \][/tex]
[tex]\[ w = 6 \][/tex]
Thus, the width of the rectangle is [tex]\( w = 6 \)[/tex] cm.
#### Step 4: Calculate the Length
Using [tex]\( w = 6 \)[/tex] cm, we find the length [tex]\( l \)[/tex]:
[tex]\[ l = w + 7 \][/tex]
[tex]\[ l = 6 + 7 \][/tex]
[tex]\[ l = 13 \][/tex] cm
#### Step 5: Calculate the Area
The area [tex]\( A \)[/tex] of the rectangle is given by:
[tex]\[ A = l \times w \][/tex]
[tex]\[ A = 13 \times 6 \][/tex]
[tex]\[ A = 78 \][/tex] square cm
#### Step 6: Find Half of the Length of the Rope
Given farone is 720 metres, we need to find half of the length of the rope:
[tex]\[ \text{Half of 720 metres} = \frac{720}{2} = 360 \][/tex] metres
### Summary
1. The width of the rectangle is [tex]\( 6 \)[/tex] cm.
2. The length of the rectangle is [tex]\( 13 \)[/tex] cm.
3. The area of the rectangle is [tex]\( 78 \)[/tex] square cm.
4. Half of the length of the rope (farone) is [tex]\( 360 \)[/tex] metres.
Let's simplify the problem given in part (b) step by step.
### Part (b): Solving the Problem
#### Step 1: Understanding the Problem
We are given a wire of length 38 cm that is bent into the shape of a rectangle. The length of the rectangle is 7 cm more than the width. We need to find:
1. The width and length of the rectangle.
2. The area of the rectangle.
3. Given farone is 720 metres, find half of the length of the rope.
#### Step 2: Representing the Variables
Let's denote:
- The width of the rectangle by [tex]\( w \)[/tex] cm.
- The length of the rectangle by [tex]\( l \)[/tex] cm.
According to the problem, the length [tex]\( l \)[/tex] is 7 cm more than the width [tex]\( w \)[/tex]:
[tex]\[ l = w + 7 \][/tex]
#### Step 3: Using Perimeter to Formulate Equations
The perimeter [tex]\( P \)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (l + w) \][/tex]
Given that the wire length (which is the perimeter) is 38 cm, we can write:
[tex]\[ 2 \times (l + w) = 38 \][/tex]
Substitute [tex]\( l = w + 7 \)[/tex] into the perimeter equation:
[tex]\[ 2 \times ((w + 7) + w) = 38 \][/tex]
Simplify and solve for [tex]\( w \)[/tex]:
[tex]\[ 2 \times (2w + 7) = 38 \][/tex]
[tex]\[ 4w + 14 = 38 \][/tex]
[tex]\[ 4w = 24 \][/tex]
[tex]\[ w = 6 \][/tex]
Thus, the width of the rectangle is [tex]\( w = 6 \)[/tex] cm.
#### Step 4: Calculate the Length
Using [tex]\( w = 6 \)[/tex] cm, we find the length [tex]\( l \)[/tex]:
[tex]\[ l = w + 7 \][/tex]
[tex]\[ l = 6 + 7 \][/tex]
[tex]\[ l = 13 \][/tex] cm
#### Step 5: Calculate the Area
The area [tex]\( A \)[/tex] of the rectangle is given by:
[tex]\[ A = l \times w \][/tex]
[tex]\[ A = 13 \times 6 \][/tex]
[tex]\[ A = 78 \][/tex] square cm
#### Step 6: Find Half of the Length of the Rope
Given farone is 720 metres, we need to find half of the length of the rope:
[tex]\[ \text{Half of 720 metres} = \frac{720}{2} = 360 \][/tex] metres
### Summary
1. The width of the rectangle is [tex]\( 6 \)[/tex] cm.
2. The length of the rectangle is [tex]\( 13 \)[/tex] cm.
3. The area of the rectangle is [tex]\( 78 \)[/tex] square cm.
4. Half of the length of the rope (farone) is [tex]\( 360 \)[/tex] metres.
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
