a Note that we have three types of tiles. One of the tiles has an area of x. Assume that its width and length is 1 and x units, respectively.
B
b Note that we have three types of tiles. One of the tiles has an area of x. Assume that its width and length is 1 and x units, respectively.
A
aProduct: (x+3)(x+1)
Sum: x^2+4x+3
B
bProduct: (x+2)(2x+1)
Sum: 2x^2+5x+2
Practice makes perfect
a First we will assume that we have three types of tiles.
In the given diagram we see three more tiles besides the labeled one with an area of x. Let's add that to the diagram.
Examining the tiles, we see that one of the rectangles has an area of x. Since the area of a rectangle is width times length, its dimension must be 1 and x. Assuming the width equals 1 unit, we can add some more information to the diagram.
Now we have all information we need to calculate the areas of the remaining tiles.
The area of the big rectangle is the product of its width and length. However, it is also the sum of all the tiles contained within it. We get two different expressions.
Product:& (x+3)(x+1)
Sum:& x^2+4x+3
b Like in Part A, we have the same three types of tiles. Let's start by writing the area on the rest of the tiles that are x square units.
Again, assuming that the width and length of the x-tile is 1 and x units, respectively, we can add the lengths alongside the big rectangle's width and length.
Now we have all information we need to calculate the areas of the remaining tiles.
The area of the big rectangle is the product of its width and length. However, it is also the sum of all the tiles contained within. We get two different expressions.
Product:& (x+2)(2x+1)
Sum:& 2x^2+5x+2
