Posters aren't complicated, right?

let length of poster = x let width of poster = y now length is 25cm more than the width therefore, x= y+25 now area of poster = x y=(y+25) y=y^2 + 25y ------1 now area of cardboard = X Y as border remains 5 cm more from each side X = x + 10 (5 from both side) and Y = y + 10 now area of cardboard = X Y = (x+10) * (y + 10) = (y + 35) * (y + 10) = y^2 + 45y +350 ----------2 now area of boarder = 1350 cm^2 also area of boarder = (area of cardboard) - (area of poster) = 1350 thus, area of boarder = (y^2 + 45y +350) - (y^2 + 25y) => y =50 thus, x = 75 cm.

Since the area of the border is 1350 cm squared, you need to get the area of the whole piece of cardboard, and then subtract that from the area of the poster to get the area of the boarder which is 1350 cm squared. Let's say w is the width. Since we know that the length of the poster is 25+w and that the cardboard is 5 cm more than the length and width in any direction, to get the length of the cardboard we just add 10 to the length and the width to get (35+w) (10+w) to get the area of the cardboard and then subtract that from (25+w) (w) =1350cm squared. Then use box method or whatever you use to multiply them to get the areas which is w squared +45w+350-(w squared+25w)=1350cm squared. Do most of the math to get 20w=1000cm squared then width is 50 cm and length is w +25 so the length is 75cm

Let's label the sides of the poster $x$ and $x+25$ . When we draw the border with length 5, we see that it creates 4 25 square inches (in each of the corners). Subtracting these squares from the total area of the border (1,350), we get 1,250 as the area of the leftover rectangles formed when the squares are removed. The are of the leftover rectangles are $5(x)+5(x)+5(x+25)+5(x+25)$ . Setting this equal to 1,250 and solving for $x$ , we see that $x=50$ . One thing about this problem is that it was unclear whether the length of the poster is the longer or shorter side. Either way, if one side is 50, the longer one is 75, which turns out to be the answer. Great problem!

Let's solve it without using variables!

First, the corners of the border account for four pieces of $5\times 5 = 25$ square cm each. Let's subtract those out. That leaves us with $1350 - 4\times 25 = 1250$ for the area of the remainder of the border.

Next, since the border is 5 cm wide every where, we divide this area by 5 to find the circumference of the poster. It is $1250 / 5 = 250$ cm.

The top and bottom are each 25 cm longer than the left and right side. This accounts for $2 \times 25 = 50$ cm of the circumference. If we shorten the poster to a square, its circumference would be $250 - 50 = 200$ cm.

A square consist of four equal sides, so we divide the circumference by four: $200 / 4 = 50$ cm is the width of the poster.

Finally, the length must be $50 + 25 = \boxed{75}$ cm.

(I know, I know--writing something like $(x-15)(x+10) - (x-25)x = 1350$ looks a lot cooler, but from that moment onward everything is a whole lot less intuitive...)

Total Board Area - Poster Area = Border Area Let poster width is 'x' so poster height will be 'x+25'. So poster area = x * (x+25) Total Board Area = (x+5+5) * (x+25+5+5) so ( (x+5+5) * (x+25+5+5) ) - ( x * (x+25) ) = 1350 On solving the equation we get x = 50. our length is x+25 i.e 75cm

say length is x+25 $\left \{ (x+25)+5+5 \right \}\times (x+5+5)-x(x+25)=1350$ $=> \left ( x^{2}+ 45x+350\right )-\left ( x^{2} +25x\right )=1350$ $=> 2x+35=135$ so, x+25=length=75