a rectangle is three times as long as it is wide. If its length and width are both increased by 3m, its area is increased by 81m2.

Question

What is it's original dimensions?

 

Answer 1

Let a be the length and b the width; a=3b

Area=A=a*b=3b²

If we replace a by a+3 and b by b+3, then

A+81=(a+3)(b+3)

3b²+81=(3b+3)(b+3)

b²+27=(b+1)(b+3)

b²+27=b²+4b+3

4b=24

b=6

a=3b=18

 

The area was 6*18=108, now (after increasing by 3 units) it is 9*21=189=108+81

 

 

 

 

 

 

Answer 2

 

   The original rectangle has area 3w * w = 3w^2 because it is 3 times as long as it is wide

 

   Adding 3 to the length and width, the area becomes:

        (3w+3)(w+3)

 

 

  The difference between these areas is 81 square meters

 

 So (3w + 3)(w+3) - 3w^2 = 81

 

      3w^2 + 9w + 3w + 9 - 3w^2 = 81

 

      12w + 9 = 81  <--- combines like terms; 3w^2 cancels out

 

     12w = 72 <--- subtracts 9 from both sides

 

       w = 6

 

  The original rectangle has width w=6 and length L=18 and has area 6 x 18 = 108

  The bigger rectangle has width w=9 and length L=21 and has area 189.

   189 - 108 = 81