a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle

Step-by-step explanation:

Let l = the original length of the original rectangle

Let w = the original width of the original rectangle

From the description of the problem, we can construct the following two equations

l=2*w (Equation #1)

(l+4)*(w+4)=l*w+88 (Equation #2)

Substitute equation #1 into equation #2

(2w+4)*(w+4)=(2w*w)+88

2w^2+4w+8w+16=2w^2+88

collect like terms on the same side of the equation

2w^2+2w^2 +12w+16-88=0

4w^2+12w-72=0

Since 4 is afactor of each term, divide both sides of the equation by 4

w^2+3w-18=0

The quadratic equation can be factored into (w+6)*(w-3)=0

Therefore w=-6 or w=3

w=-6 can be rejected because the length of a rectangle can't be negative so

w=3 and from equation #1 l=2*w=2*3=6

I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.

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3 years ago