Question

Darrin is hanging 200 feet of Christmas garland on the fencing that encloses his rectangular front yard. The length is 5 feet less than 5 times the width. Find the length and width of the fencing that encloses his rectangular front yard.

Answer 1

Width = 17.5 feet and length= 82.5 feet

Explanation:

Let w be the width of the rectangular front yard.

It is given that the length is 5 feet less than 5 times the width.

`l=5w-5`

The perimeter of a rectangle is

`Perimeter=2(l+w)`

Perimeter of rectangular front yard is

`Perimeter=2(5w-5+w)`

`Perimeter=2(6w-5)`

`Perimeter=12w-10` .... (1)

Darrin is hanging 200 feet of Christmas garland on the fencing that encloses his rectangular front yard.

`Perimeter=200` .... (2)

From (1) and (2) we get

`12w-10=200`

Add 10 on both sides.

`12w=200+10`

`12w=210`

Divide both sides by 12.

`w=(210)/(12)`

`w=17.5`

Therefore, the width of the rectangular yard is 17.5 feet.

`l=5w-5`

Substitute w=17.5.

`l=5(17.5)-5`

`l=87.5-5`

`l=82.5`

Therefore, the length of the rectangular yard is 82.5 feet.

Answer 2

Width = 17.5 feet,

Length = 82.5 feet.

Explanation:

We have been given that Darrin is hanging 200 feet of Christmas garland on the fencing that encloses his rectangular front yard. This means that perimeter of rectangular front yard is 200 feet.

We know that perimeter of rectangle is two times the sum of length and width of rectangle. We can represent this information in an equation as:

`2(l+w)=200...(1)`

We are also told that the length is 5 feet less than 5 times the width. We can represent this information in an equation as:

`l=5w-5...(2)`

Upon substituting equation (2) in equation (1), we will get:

`2(5w-5+w)=200`

`2(6w-5)=200`

`12w-10=200`

`12w-10+10=200+10`

`12w=210`

`(12w)/(12)=(210)/(12)`

`w=17.5`

Therefore, the width of rectangular front is 17.5 feet.

Upon substituting
`w=17.5` in equation (2), we will get:

`l=5(17.5)-5`

`l=87.5-5`

`l=82.5`

Therefore, the length of rectangular front is 82.5 feet.