Chapter 11.1, Problem 56AYU

To determine

To find: The dimensions of a rectangular field if the difference between its length and width is 50 meters and the length of fence required to enclose the field is 3000 meters.

Answer to Problem 56AYU

Solution:

Length $=775$ meters; Width $=725$ meters.

Explanation of Solution

Given:

Difference between the length and width of the rectangular field is 50 meters.

Length of fence to enclose the field $=3000$ meters.

Calculation:

Let the width of the rectangular field be $x$ meters.

Then the length of the field is $=(x+50)$ meters.

Length of fence to enclose the field = Perimeter of the rectangular field $=3000$ meters.

We know that in a rectangle,

Perimeter $=2\times ($ length $+$width)

Thus we get, $2\times (x+50+x)=3000$

  $2\times (2x+50)=3000$$4x+100=3000$

  $4x=3000-100$

  $4x=2900$

  $x=\frac{2900}{4}$

  $x=725$

Therefore the width of the rectangular field is 725 meters.

Then the length of the rectangular field becomes = $725+50=775$ meters.

Hence the dimensions, that is, the length and width of the rectangular field are 775 meters and 725 meters respectively.