SOLUTION: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet then the area is multiplied by 3. What was the size of the original rectangle?

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet then the area is multiplied by 3. What was the size of the original rectangle?       Log On


   

Question 1166186: The length of a rectangle is 10 feet longer than it is wide. If each side is increased 10 feet then the area is multiplied by 3. What was the size of the original rectangle?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Dimensions w and w+10
Area w(w+10)

%28w%2B10%29%28w%2B20%29=3%28w%5E2%2B10w%29
Simplify and solve,...

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let the width be w feet;

Then the length  is (w+10) feet


The condition says

    (w+10+10)*(w+10) = 3w(w+10).


At this point, do not make FOIL.

Do not create a quadratic equation --- there is a simpler way.

Cancel the common factor (w+10) in both sides.  You will get then


    w + 10 + 10 = 3w

    w + 20      = 3w

    3w - w = 20.

    2w     = 20,

     w     = 10.


So, the dimensions of the original rectangle are  10 ft (the width)  and  10+10 = 20 ft (the length).

Solved.

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You may also think,  what the condition means,  geometrically.

Thinking in this way,  you will be able to solve the entire problem  MENTALLY  (in your head).