Question

a rectangular field is 50 yards wide and 100 yards long. give the length and width of another rectangular field that has the same perimeter but a smaller area. width = □ yards length = □ yards

Explanation:

Step1: Calculate the perimeter of the original rectangle

The formula for the perimeter of a rectangle is \( P = 2\times (length + width) \). For the original rectangle with width \( w = 50 \) yards and length \( l = 100 \) yards, we have:
\( P = 2\times(100 + 50) = 2\times150 = 300 \) yards.

Step2: Let the new width be \( x \) and find the new length

Let the width of the new rectangle be \( x \) yards. Since the perimeter is 300 yards, using the perimeter formula \( P = 2\times (l + x) \), we can solve for the length \( l \):
\( 300 = 2\times(l + x) \)
Divide both sides by 2: \( 150 = l + x \), so \( l = 150 - x \)

Step3: Calculate the area of the original rectangle and find a new rectangle with smaller area

The area of the original rectangle is \( A = length\times width = 100\times50 = 5000 \) square yards.
We want a rectangle with area \( A' = l\times x=(150 - x)\times x = 150x - x^{2} \) which is less than 5000.
Let's try a width smaller than 50, say \( x = 40 \) yards. Then the length \( l = 150 - 40 = 110 \) yards.
The area of this new rectangle is \( 40\times110 = 4400 \) square yards, which is less than 5000. (We can also try other values, for example, width = 30, length = 120, area = 3600, etc. Here we choose one possible pair.)

Answer:

width = 40 yards
length = 110 yards

(Note: There are multiple correct answers, for example, width = 30, length = 120; width = 20, length = 130; etc. The above is just one possible solution.)