The length of a rectangle is 8 feet longer than it is wide. If each side is increased 8 feet, then the area is multiplied by 4. What was the size of the origina

Question

The length of a rectangle is 8 feet longer than it is wide. If each side is increased 8 feet, then the area is multiplied by 4. What was the size of the original rectangle?

Answer 1

So your length is 8 more than the width:

Width=x then Length=x+8

Increase the length by 8 and the width by eight so you get a new rectangle of dimentions W=x+8 and L=x+8+8=x+16

the area of the second rectangle is 4 times nore than the first which means (x+8)(x+16)=4(x)(×+8) => x^(2) +24x+128=4x^(2)+32x => 3x^(2)+8x-128=0 which gives you (3x-16)(x+8)=0 so x+8=0 => x=-8 which is imposible since x is the width which can not be negative so we are left with 3x-16=0 => x=16/3. So go back to the original rectangle  w=16/3ft and L=(16/3 +8) ft

Multiply for the area if needed.

I hope that was helpful!

Answer 2

w=width

w+8=length

A=w(w+8)

A=w^2+8w

 

increase the sides

width=w+8

length=w+8+8=w+16

A=(w+8)(w+16)

A=w^2+24w+128

w^2+24W+128=4(w^2+8w)

w^2+24w+128=4w^2+32w

4w^2+32w-w^2-24w-128=0

3w^2+8w-128=0

factor

(3w-16)(w+8)=0

3w-16=0

3w=16

w=16/3 feet or 5 1/3 feet

l=13 1/3 feet

5 1/3 feet and 13 1/3 feet are the original measurements

check: (16/3)(40/3)=640/9

13 1/3 * 21 1/3=(40/3)(64/3)=2560/9  (2560 is 4*640 or you should see that in the two products that 64 is 4*16)