SOLUTION: a rectangle is 5 ft longer than its width. if the lenght is shortened by 2 ft and the width is increased by 1 ft, the area remains unchanged. find the area.
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Question 213072: a rectangle is 5 ft longer than its width. if the lenght is shortened by 2 ft and the width is increased by 1 ft, the area remains unchanged. find the area. Answer by aggie_tutor(13) (Show Source):
You can put this solution on YOUR website! They trick you a little: They are asking for the area, but first you have to solve for the actual width. Fortunately this works out pretty easily. Let width above be w and rectangles A and B.
Area of rectangle A:
Width = w
Length = w + 5
Area rectangle A: (Length * Widgth) AreaA = (w)*(w+5)
Area of rectangle B:
Width = w + 1
Length = (w + 5) - 2
Area rectangel B: (Length * Width) AreaB =
Now the problem says the two areas are equal. Set each area computation equal to the other, and solve for width w.
AreaA = AreaB
Simplify with distributive property:
Combine like terms.
Subtrace the w^2 from each side to get rid of it. The square term cancels itself out, which is really convenient so we don't have to take a root of anything (!).
Subtract 4w from both sides to isolate the number 3.
Then plug in w to one of the rectangles to get the area.
AreaA = Length * Width =
Check your answer by subbing w in the other rectangle:
AreaB = Length * Width =
AreaA = AreaB !
