Problem 10, Part A
The larger outer square is x cm by x cm, where x is some placeholder for a positive number. For example, if x = 10, then we have a 10 cm by 10 cm square.
Unfortunately, we don't know what number replaces x.
What we can determine is the connection between the larger square and the smaller one. The frame is 3 cm wide in all directions (aka it's a uniform width). What this allows for is that the smaller square is x-6 by x-6. The 6 is from subtracting two copies of 3, so ultimately we're subtracting off 2*3 = 6 to form the smaller dimensions.
You can think of it like x-3-3 = x-6
So we need to expand out (x-6)(x-6) like so
(x-6)(x-6)
y(x-6)
xy - 6y
x(x-6) - 6(x-6)
x^2-6x - 6x+36
x^2-12x+36
In the second step, I replaced one copy of (x-6) with y, after which I used the distributive rule. After distributing, I plugged in y = x-6 and distributed again. The last step is combining like terms. As an alternative, you can use either the box method or the FOIL rule to find the answer.
The expression x^2-12x+36 is the expanded form of the area of the smaller square, which is the picture itself without the frame involved.
x^2 - 12x + 36
Problem 10, Part B
We follow the same idea as the previous part. This time, we have a rectangle that isn't a square.
Along the horizontal, the (x+30) will drop to (x+30)-5-5 = x+20 when we consider just the picture without the frame. In other words, the smaller inner rectangle has a horizontal dimension of (x+20) cm.
The vertical component of this smaller rectangle is x-5-5 = x-10 cm
So,
area = length*width
area = (horizontal dimension)*(vertical dimension)
area = (x+20)(x-10)
area = y(x-10) let y = x+20
area = xy - 10y distribute
area = x(x+20) - 10(x+20) ... plug in y = x+20
area = x^2+20x - 10x - 200 distribute again
area = x^2 + 10x - 200
x^2 + 10x - 200
Side note: the units for this area, and for part A, are in square cm.
