Question 1097200
You need more help than that.
 
You are half right on question 5a.
As you draw each new square,
the new square area is half of the area of the previous square.
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
red(rectangle(-1,-1,1,1)),
green(triangle(-1,0,0,1,0,-1)),
green(triangle(1,0,-1,0,0,-1)),
line(0,-1,1,0),line(1,0,0,1),
line(0,1,-1,0),line(-1,0,0,-1)
)}}}
The square can be split into 4 more triangles,
as I did with those green lines,
and the other half of the previous square area is made of the 4 new triangles.
The areas of squares and triangles form a geometric sequence with common ratio 1/2
So, the areas are 32,16,8,4,2,1,1/2,1/4, and so on.
The terms of that sequence are called {{{A[n]}}}
The second blank (area of the third triangle) is indeed {{{A[3]=8}}} .
 
Because areas are made by multiplying lengths of sides and a convenient factor,
lengths of sides form a geometric sequence with common ratio
{{{sqrt(1/2)=sqrt(2)/2}}} .
So, the sides are 8,4sqrt(2),4,2sqrt(2),2,sqrt(2),1, and so on.
The terms of that sequence of lengths are called {{{X[n]}}} .
The first blank (side of the second triangle) is not 6, but {{{X[2]=4sqrt(2)}}} .
 
5b. You may have misinterpreted that question.
The question asks for {{{A[6]}}} , the 6th term of the sequence of areas that I listed above.
I do not know how you came up with your answer, but the sequence of areas is
32,16,8,4,2,1,1/2,1/4, and so on, and the 6th term there is
{{{A[6]=1}}} .
 
5c. You did not include a picture so that we could figure out what area was shaded in the picture that came with your question.
However, I did an internet search using a phrase with your question.
I found the question to show one shaded triangle from the each step,
all adjacent to the triangles in the steps before and after,
forming a clockwise spiral.
The area shaded after an infinite number of steps is the sum of the {{{A[n]}}} terms from start to infinity:
32+16+8+4+2+1+....=64.
If you had to "show your work" the teacher may expect you to say that
the areas form a geometric sequence with common ratio {{{r=1/2}}} .
and because that ratio is less than 1, the series (the sum) converges to
{{{S=A[1]/(1-r)}}}{{{"="}}}{{{32/(1-1/2)}}}{{{"="}}}{{{32/(1/2)}}}{{{"="}}}{{{32*2=64}}} .
However, a fifth grader (or someone smarter than that)
would realize that each sum gets closer to 64,
and that what it needs to get to 64 is exactly the last term added.
Adding we would never get to 64,
because we would need to add infinite terms,
and our lives are finite,
but the sum to infinity is 64,
because we see that {{{A[infinity]}}} ,
if there was such a real number, would be zero.
If the first square has side length {{{k}}} ,
the area of that square would be {{{k^2}}} and the area of each of the first set of tr triangles would be
{{{A[1]=k^2/8}}} .
Then the sum of infinite triangles would be {{{2A[1]=k^2/4}}} .
If that is equal to {{{k}}} ,
{{{k^2/4=k}}} , meaning that if {{{k<>0}}} ,
and there really was an initial square,
we can divide both sides by {{{k}}} to get
{{{k/4=1}}} and {{{highlight(k=4)}}} .