Chapter 4.9, Problem 4P

To determine

To calculate:need to find the dimension of the poster

Answer to Problem 4P

Width of the poster is 50 cm and Length of the poster is 75 cm

Explanation of Solution

Given information: Length of poster is 25 cm taller than its width.

Calculation: Let $w$ be the width of poster, then $\left(w+25\right)$ is the length of poster. Then its area is length * breadth, i.e.

  $A_B=w\left(w+25\right)$

  $A_B=w^2+25w\mathrm{...}\left(i\right)$

It is mounted on a piece of cardboard so that there is a 5 cm border on all sides, so its length and width will increase by 5+5 cm each, so length of poster and border is $w+5+5=w+10$ cm and width of poster and border is $w+25+5+5=w+35$ cm. Then area is

  $A_{P+B}=\left(w+10\right)\left(w+25+10\right)$

  $A_{P+B}=\left(w+10\right)\left(w+35\right)$

  $A_{P+B}=w^2+10w+35w+350$

  $A_{P+B}=w^2+45w+\mathrm{350...}\left(ii\right)$

Area of border alone is given as $1350c{m}^2$ .

Area of border alone = Area of border and poster - Area of poster , i.e.

  $A_B=A_{P+B}-A_P$

  $1350=w^2+45w+350-\left(w^2+25w\right)$

  $1350=20w+350$

  $20w=1000$

  $w=50$

Width of the poster is $w=50$ cm and length of the poster is $w+25=50+25=75$ cm.

Conclusion: Width of the poster is 50 cm and Length of the poster is 75 cm.