Chapter 4.9, Problem 1P

To determine

To calculate:Find the original dimension and make a sketch

Answer to Problem 1P

Length and width of original rectangle are 15 cm and 5 cm.

Explanation of Solution

Given information: length of rectangle is 3 times of its width and for new rectangle both length and width are decreased by 2 cm.

Calculation: Let $w$ be the width of rectangle, then its length will be $3w$ .

Now according to the given information, the width and length of new rectangle are $w-2$ and $3w-2$ respectively.

Now we have length and width of both rectangles, so we can make its sketch

  

Area of a rectangle is $A=l*w$ ,so

Area of original rectangle is $A_{or}=w*3w=3w^2$

Area of new rectangle is $A_{nr}=\left(w-2\right)*\left(3w-2\right)$

  $A_{nr}=3w^2-2w-6w+4$

  $A_{nr}=3w^2-8w+4$

according to question, Area of new rectangle is $36c{m}^2$ less than the Area of original rectangle, so

  $3w^2-\left(3w^2-8w+4\right)=36$

  $8w-4=36$

  $8w=40$

  $w=5$

and $3w=3*5=15$ ,so

Width of original rectangle is 5 cm and its length is 15 cm.

Conclusion: Length and width of original rectangle are 15 cm and 5 cm.