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

$\text{Length }=\text{ 775 meters};\text{ Width }=\text{ 72}5\text{ meters}$.

Explanation of Solution

Given:

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

Length of fence to enclose the field $=\text{ 3000 meters}$.

Calculation:

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

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

Length of fence to enclose the field $=$ Perimeter of the rectangular field $=\text{ 3000 meters}$.

We know that in a rectangle,

$\text{Perimeter }=2\times \left(\text{length }+\text{ width}\right)$

Thus we get,

$2\times \left(x+50+x\right)=3000$

$2\times \left(2x+50\right)=3000$

$4x+100=3000$

$4x=3000-100$

$4x=2900$

$x=\frac{2900}{4}$

$x=725$

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.