Alright, let's go through each part step-by-step.
### Part 1: Drawing the Rectangles
1. First Rectangle:
- Length = 9 cm
- Breadth = 5 cm
- Area Calculation:
[tex]\[
\text{Area}_1 = \text{Length}_1 \times \text{Breadth}_1 = 9 \, \text{cm} \times 5 \, \text{cm} = 45 \, \text{cm}^2
\][/tex]
2. Second Rectangle:
- Given Area (same as the first rectangle): [tex]\( \text{Area}_1 = 45 \, \text{cm}^2 \)[/tex]
- New Length = 15 cm
- Finding the New Breadth:
For the area to remain the same, we use the same area formula:
[tex]\[
\text{Area}_2 = \text{New Length} \times \text{Breadth}_2
\][/tex]
Since [tex]\(\text{Area}_2 = 45 \, \text{cm}^2 \)[/tex] and [tex]\(\text{New Length} = 15 \, \text{cm} \)[/tex]:
[tex]\[
45 \, \text{cm}^2 = 15 \, \text{cm} \times \text{Breadth}_2
\][/tex]
[tex]\[
\text{Breadth}_2 = \frac{45 \, \text{cm}^2}{15 \, \text{cm}} = 3 \, \text{cm}
\][/tex]
3. Third Rectangle:
- Given Area (same as the first rectangle): [tex]\( \text{Area}_1 = 45 \, \text{cm}^2 \)[/tex]
- New Length = 45 cm
- Finding the New Breadth:
Similarly, using the area formula:
[tex]\[
45 \, \text{cm}^2 = 45 \, \text{cm} \times \text{Breadth}_3
\][/tex]
[tex]\[
\text{Breadth}_3 = \frac{45 \, \text{cm}^2}{45 \, \text{cm}} = 1 \, \text{cm}
\][/tex]
### Observations
- First Rectangle: 9 cm (length) and 5 cm (breadth)
- Second Rectangle: 15 cm (length) and 3 cm (breadth)
- Third Rectangle: 45 cm (length) and 1 cm (breadth)
Observation: The area of the rectangles remains constant at [tex]\(45 \, \text{cm}^2\)[/tex]. However, as the length increases, the breadth decreases proportionally to ensure that the product of length and breadth (i.e., the area) remains the same.
### Part 2: Algebraic Expressions
(a) Finding the monomial:
- Given: The monomial divided by [tex]\(3x^2\)[/tex] gives [tex]\(9xy\)[/tex].
- Let the monomial be [tex]\(m\)[/tex].
- Form the equation:
[tex]\[
\frac{m}{3x^2} = 9xy
\][/tex]
[tex]\[
m = 9xy \times 3x^2
\][/tex]
[tex]\[
m = 27x^3y
\][/tex]
- So, the monomial is [tex]\(27x^3y\)[/tex].
(b) Finding the binomial:
- Given: The binomial divided by [tex]\(3x^2\)[/tex] gives [tex]\(9xv\)[/tex].
- Let the binomial be [tex]\(b\)[/tex].
- Form the equation:
[tex]\[
\frac{b}{3x^2} = 9xv
\][/tex]
[tex]\[
b = 9xv \times 3x^2
\][/tex]
[tex]\[
b = 27x^3v
\][/tex]
- So, the binomial is [tex]\(27x^3v\)[/tex].
### Summary
- Areas:
- First Rectangle: [tex]\(45 \, \text{cm}^2\)[/tex]
- Second Rectangle: Length = 15 cm, Breadth = 3 cm
- Third Rectangle: Length = 45 cm, Breadth = 1 cm
- Algebraic Expressions:
- Monomial: [tex]\(27x^3y\)[/tex]
- Binomial: [tex]\(27x^3v\)[/tex]
This systematic process ensures that the area calculations and algebraic derivations are clear and accurate.
