Mr. James wanted to plant a garden that would bein the shape of arectangle. He was given 80 feet of fencing toenclose his garden. Hewants the length to be 10 feet more thantwice the width. What are the dimensions, in feet, for a rectangular garden that will use exactly 80 feet offencing?

Hayley Bernard

Hayley Bernard

Answered question

2022-07-27

Mr. James wanted to plant a garden that would bein the shape of arectangle. He was given 80 feet of fencing toenclose his garden. Hewants the length to be 10 feet more thantwice the width. What are the dimensions, in feet, for a rectangular garden that will use exactly 80 feet offencing?

Answer & Explanation

escobamesmo

escobamesmo

Beginner2022-07-28Added 18 answers

let the width of the rectangular garden = x feet.
as Mr. James wants the length to be 10 feet more twice the width,hence the length is=(2x+10)feet.
hence the perimeter of the garden is
= 2*[x + (2*x + 10)]feet
= 2*[3x + 10] feet
= (6x + 20)feet
he was given 80 feet of fencing to enclose hisgarden.
so
6x + 20 = 80
=> 6x = 80 - 20 = 60
=> x = 60/6 = 10
so, width = x = 10 feet.
hence length =2x + 10 = 2*10+10 =30feet.
answer: length = 30 feet;
width = 10 feet;
Urijah Estes

Urijah Estes

Beginner2022-07-29Added 5 answers

Set length = l and width = w
l = 2w + 10
80 = 2l + 2w *l and w are multiplied by 2 because we're finding the perimeter
Substitute in l in terms of w to get the new equation 80 = 2(2w +10) + 2w
Solve for w, which will equal 10
Plug back into length equation l = 2w + 10 to get l = 30
Now find the perimeter: 2l + 2w = 2(30) + 2(10) =80 *checks out!
The garden will be 30 ft long by 10 ft wide

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