Question

the ratio of the area of a square to the area of a rectangle is 2:3. The ratio of the length and width of the rectangle is 3:2.
Find the perimeter of the square,if the perimeter of the rectangle is 80 cm?

Answer 1

Let's denote the side length of the square as \( s \) and the length and width of the rectangle as \( l \) and \( w \), respectively.

The area of the square is:
\[
\text{Area of square} = s^2
\]

The area of the rectangle is:
\[
\text{Area of rectangle} = l \times w
\]

According to the problem, the ratio of the area of the square to the area of the rectangle is \( 2:3 \), which can be written as:
\[
\frac{s^2}{l \times w} = \frac{2}{3}
\]
This implies:
\[
3s^2 = 2(l \times w) \quad (1)
\]

The problem also states that the ratio of the length and width of the rectangle is \( 3:2 \). We can express \( l \) and \( w \) in terms of a variable \( x \):
\[
l = 3x \quad \text{and} \quad w = 2x
\]

Now, we can find the perimeter of the rectangle, which is given as \( 80 \) cm:
\[
\text{Perimeter of rectangle} = 2(l + w) = 2(3x + 2x) = 2(5x) = 10x
\]
Setting this equal to \( 80 \):
\[
10x = 80 \quad \Rightarrow \quad x = 8
\]

Now we can find the dimensions of the rectangle:
\[
l = 3x = 3(8) = 24 \quad \text{and} \quad w = 2x = 2(8) = 16
\]

Next, we can calculate the area of the rectangle:
\[
\text{Area of rectangle} = l \times w = 24 \times 16 = 384
\]

Now, substituting this area into equation (1):
\[
3s^2 = 2(384)
\]
\[
3s^2 = 768
\]
\[
s^2 = \frac{768}{3} = 256
\]
\[
s = \sqrt{256} = 16
\]

Finally, we can find the perimeter of the square:
\[
\text{Perimeter of square} = 4s = 4 \times 16 = 64 \text{ cm}
\]

Thus, the perimeter of the square is:
\[
\boxed{64 \text{ cm}}
\]