the rectangular field?
5) A rectangular field has the area equal to that of a square field of side 40 m.
rectangular field is 30 m,
(i) Find the area of the rectangular field.
(ii) Find the length of the rectangular field.
of the
(iii) If the length and width of the rectangular field are doubled then what would be the area of
the rectangular field?
ectively



Answer :

Alright, let's solve the problem step-by-step.

### Given:
- A square field with a side of 40 meters.
- A rectangular field with a width of 30 meters and the same area as the square field.

### (i) Find the area of the rectangular field

First, we need to find the area of the square field.

The area of a square is given by the formula:
[tex]\[ \text{Area}_{\text{square}} = \text{side}^2 \][/tex]

For the square field:
[tex]\[ \text{Area}_{\text{square}} = 40 \, \text{m} \times 40 \, \text{m} = 1600 \, \text{m}^2 \][/tex]

Since the rectangular field has the same area as the square field:
[tex]\[ \text{Area}_{\text{rectangular}} = 1600 \, \text{m}^2 \][/tex]

### (ii) Find the length of the rectangular field

We are given the width of the rectangular field as 30 meters. To find the length, we use the area of the rectangular field and the width.

The area of a rectangle is given by:
[tex]\[ \text{Area}_{\text{rectangular}} = \text{length} \times \text{width} \][/tex]

We can rearrange this formula to solve for the length:
[tex]\[ \text{length} = \frac{\text{Area}_{\text{rectangular}}}{\text{width}} \][/tex]

Plugging in the values:
[tex]\[ \text{length} = \frac{1600 \, \text{m}^2}{30 \, \text{m}} = \frac{1600}{30} \, \text{m} = \frac{160}{3} \, \text{m} \approx 53.33 \, \text{m} \][/tex]

### (iii) If the length and width of the rectangular field are doubled, what would be the area of the rectangular field?

If both the length and width of the rectangular field are doubled:
[tex]\[ \text{new length} = 2 \times \left( \frac{160}{3} \right) \, \text{m} = \frac{320}{3} \, \text{m} \][/tex]

[tex]\[ \text{new width} = 2 \times 30 \, \text{m} = 60 \, \text{m} \][/tex]

Next, we find the area of the new, larger rectangular field:
[tex]\[ \text{Area}_{\text{new rectangular}} = \text{new length} \times \text{new width} \][/tex]

Plug in the values:
[tex]\[ \text{Area}_{\text{new rectangular}} = \left( \frac{320}{3} \, \text{m} \right) \times 60 \, \text{m} = \frac{320 \times 60}{3} \, \text{m}^2 = \frac{19200}{3} \, \text{m}^2 = 6400 \, \text{m}^2 \][/tex]

### Summary

- (i) The area of the rectangular field is 1600 square meters.
- (ii) The length of the rectangular field is approximately 53.33 meters ([tex]\( \frac{160}{3} \)[/tex] meters).
- (iii) If the length and width of the rectangular field are doubled, the area of the new larger rectangular field would be 6400 square meters.