Let's go through each part of the question step-by-step.
### Question 5: Rectangular Field Problem
Given:
- The width of the rectangular field is 50 metres.
- The total length of the required fencing is 260 metres.
- We need to determine the correct equation for [tex]\( x \)[/tex] which represents the length of the field.
Options Given:
- A. [tex]\( x + 100 = 260 \)[/tex]
- C. [tex]\( 4x + 200 = 260 \)[/tex]
- B. [tex]\( 2x + 50 = 260 \)[/tex]
- D. [tex]\( 2x + 100 = 260 \)[/tex]
Solution:
To determine which equation is correct, we need to use the given information:
- The perimeter of a rectangle is given by [tex]\( 2 \times \text{length} + 2 \times \text{width} \)[/tex].
- In this problem, the width is 50 meters, and the length is [tex]\( x \)[/tex] meters.
So the perimeter equation will be:
[tex]\[ 2x + 2 \times 50 = 260 \][/tex]
which simplifies to:
[tex]\[ 2x + 100 = 260 \][/tex]
Thus, the correct option is:
Option D: [tex]\( 2x + 100 = 260 \)[/tex]
### Question 6: Area of a Circular Tray
Given:
- The diameter of the circular tray is 28 cm.
- We need to use [tex]\(\pi = \frac{22}{7}\)[/tex].
Solution:
To find the area of the circular tray, we use the formula for the area of a circle:
[tex]\[ \text{Area} = \pi r^2 \][/tex]
First, we find the radius [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\text{diameter}}{2} = \frac{28}{2} = 14 \text{ cm} \][/tex]
Now we plug in the radius and the given value of [tex]\(\pi\)[/tex]:
[tex]\[ \text{Area} = \pi \times (14)^2 \][/tex]
[tex]\[ \text{Area} = \frac{22}{7} \times 14 \times 14 \][/tex]
[tex]\[ \text{Area} = \frac{22}{7} \times 196 \][/tex]
[tex]\[ \text{Area} = 22 \times 28 \][/tex]
[tex]\[ \text{Area} = 616 \text{ cm}^2 \][/tex]
Thus, the correct option is:
Option D: 616 cm[tex]\(^2\)[/tex]
